H C VERMA Solutions for Class 12-science Physics Chapter 12 - Magnetic Field

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Chapter 12 - Magnetic Field Exercise 230

Question 1

An alpha particle is projected vertically upward with a speed of  H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkm/s in a region where a magnetic field of magnitude 1.0T exists in the direction south to north. Find the magnetic force that acts on the  H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldparticle.

Solution 1

 H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN towards west.

Question 2

An electron is projected horizontally with a kinetic energy of 10 keV. A magnetic field of strength H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT exists in the vertically upward direction. (a) Will the electron deflect towards right or towards left of its motion? (b) Calculate the sideways deflection of the electron in travelling through 1m. Make appropriate approximations.

Solution 2

(a) By Right hand rule the electron will be deflected towards left.

(b) Kinetic energy, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (horizontally)

Magnetic force, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (along west direction)

Time taken to travel 1m in horizontal direction

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

In this time, deflection of electron in west direction

deflectionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflection=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflection=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflection=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflectionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

deflectionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

deflectionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Question 3

A magnetic field of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT exerts a force of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN on a particle having a charge of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldC and going in the X-Y plane. Find the velocity of the particle.

Solution 3

Let velocity of the particle in X-Y plane H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Force on particle, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Comparing,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Velocity, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 4

An experiment's diary reads as follows:" a charged particle is projected in a magnetic field of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT. The acceleration of the particle is found to be H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field. The number to the left H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field in the last expression was not readable. What can this number be?

Solution 4

Since, force exerted by magnetic field is always perpendicular to each other.

So, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

So, acceleration of particle is H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field.

Question 5

A 10g bullet having a charge of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field is fired at a speed of 270m/s in a horizontal direction. A vertical magnetic field of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field exists in the space. Find the deflection of the bullet due to the magnetic field as it travels through 100m. Make appropriate approximations.

Solution 5

Time to travel 100m distance horizontally,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldsec

Force on the bullet

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Deflection due to the magnetic fieldH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflectionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflection=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

deflectionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Question 6

When a proton is released from rest in a room, its starts with an initial acceleration a0 towards west. When it is projected towards north with a speed v0, it moves with an initial acceleration 3a0 towards west. Find the electric field and the maximum possible magnetic field in the room.

Solution 6

Magnetic field does not exert any force on a particle at rest.

So, initially force have been acted by electric field.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (along west)

When projected towards north, force will be exerted by both fields and in same direction as acceleration becomes thrice in same direction (along west).

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (into the paper perpendicularly)

Question 7

Consider a 10cm long portion of a straight wire carrying a current of 10A placed in a magnetic field of 0.1T making an angle of 53° with the wire. What magnetic force does the wire experience?

Solution 7

Force=i(l × B) 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

F 0.08N actin perpendicular to both wire and magnetic field.

Question 8

A current of 2A enters at the corner d of a square frame abcd of side 20cm and leaves at the opposite corner b. A magnetic field B=0.1T exists in the space in a direction perpendicular to the plane of the frame as shown in figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 8

The current will equally be divided at junction d.

So, current along dcb wire and dab wire is 1A.

Since, current, length and magnetic field for all wires is same then magnitude of force is equal.

Magnitude of force, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN

Direction of force can be seen by Right hand rule.

On da and cb wire is towards left.

On dc and ab wire is downward.

Chapter 12 - Magnetic Field Exercise 231

Question 9

A magnetic field of strength 1.0T is produced by a strong electromagnet in a cylindrical region of radius 4.0cm as shown in figure. A wire, carrying a current of 2.0A, is placed perpendicular to and intersecting the axis of the cylindrical region. Find the magnitude of the force acting on the wire.

 

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Solution 9

Length of the wire in magnetic field=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Force of wire, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN

Question 10

A wire of length l carries a current i along the X-axis. A magnetic field exists which is given as H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT. Find the magnitude of the magnetic force acting on the wire.

Solution 10

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 11

A current of 5.0A exists in the circuit shown in figure. The wire PQ has a length of 50cm and the magnetic field in which it is immersed has a magnitude of 0.20T. Find the magnetic force acting on the wire PQ.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 11

Force, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN upward in the plane of paper.

Question 12

A circular loop of radius a, carrying a current I, is placed in a two-dimensional magnetic field. The center of the loop coincides with the center of the field. The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 12

Magnetic field is radially outwards.

Magnetic field and length are perpendicular to each other.

Now, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 13

A hypothetical magnetic field existing in a region is given by H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field, where H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field denotes the unit vector along the radial direction. A circular loop of radius a, carrying a current i, is placed with its plane parallel to the X-Y plane and the center at (0, 0, d). Find the magnitude of the magnetic force acting on the loop.

Solution 13

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Magnetic force due to B0cosѲ field is radially outward which will be counter balanced by each other. So, force by B0cosѲ field is zero.

Magnetic force due to B0cosѲ field is in downward direction.

So, magnetic force F = Bill SinѲ

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 14

A rectangular wire-loop of width a is suspended from the insulted pan of a spring balance as shown in figure. A current i exists in the anticlockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Solution 14

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Initially, when current is flowing in anticlockwise direction the force on AB and CD wire are counter-balanced while force on BC is in upward direction.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

So, initially TensionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Later, when current is flowing in clockwise direction the force will act in downward direction of BC wire.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

So, finally TensionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Change in TensionH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 15

A current loop of arbitrary shape lies in a uniform magnetic field B. Show that the net magnetic force acting on the loop is zero.

Solution 15

Let the arbitrary loop be rectangle PQRS having length 'l' and breadth 'b'.

Magnetic field B is perpendicular to plane of loop.

Magnitude of force on QP and SR wire isH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Both are in opposite direction. So will cancel each other.

Magnitude of force on PS and RQ wire is = Bib

Both are in opposite direction. So will cancel each other.

Therefore, net force acting on the loop is zero.

Question 16

Prove that the force acting on a current-carrying wire, joining two fixed points a and b in a uniform magnetic field, is independent of the shape of the wire.

Solution 16

Let us consider a semi-circular wire of radius R is placed in uniform magnetic field directed perpendicular of wire.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Considering two element at an angle H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field as shown in figure.

Force acting on each element H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Resultant force on both element H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Resultant forceH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (distance between initial and final point)

Question 17

A semi-circular wire of radius 5.0cm and carries a current of 5.0A. A magnetic field B of magnitude 0.50T exists along the perpendicular to the plane of the wire. Find the magnitude of the magnetic force acting on the wire.

Solution 17

Length of wireH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Magnetic field, B=0.5T

Current, i=5A

ForceH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

F=0.25N

Question 18

A wire, carrying a current i, is kept in the X-Y plane along the curve H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field. A magnetic field B exists in the z-direction. Find the magnitude of the magnetic force on the portion of the wire between x=0 and x=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field.

Solution 18

Magnetic force depends only upon distance between initial and final point not on shape of wire.

Length of wireH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Magnetic force, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 19

A rigid wire consists of a semicircular portion of radius R and two straight sections. The wire is partially immersed in a perpendicular magnetic field B as shown in figure. Find the magnetic force on the wire if it carries a current i.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 19

Magnetic force depends only upon distance between initial and final point of wire.

Length of wire=2R

Force=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

F=2iRB

Question 20

A straight, horizontal wire of mass 10mg and length 1.0m carries a current of 2.0A. What minimum magnetic field B should be applied in the region so that the magnetic force on the wire may balance its weight?

Solution 20

Under equilibrium

Magnetic force=Gravitational force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT

Question 21

Figure shows a rod PQ of length 20.0cm and mass 200g suspended through a fixed point O by two threads of lengths 20.0cm each. A magnetic field of strength 0.500T exists in the vicinity of the wire PQ as shown in figure. The wires connecting PQ with the battery are loose and exert no force on PQ. (a) Find the tension in the threads when the switch S is open. (b) A current of 2.0A is established when the switch S is closed. Find the tension in the threads now.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 21

(a) Switch 'S' is open

Weight of the rod is balanced by the tension in threads

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN

(b) When switch is closed, magnetic force acts in downward direction given by right hand rule.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

T=1.25N

Chapter 12 - Magnetic Field Exercise 232

Question 22

Two metal strips, each of length l, are clamped parallel to each other on a horizontal floor with a separation b between them. A wire of mass m lies on them perpendicularly as shown in figure. A vertically upward magnetic field of strength B exists in the space. The metal strips are smooth but the coefficient of friction between the wire and the floor is μ. A current i is established when the switch S is closed at the instant t=0. Discuss the motion of the wire after the switch is closed. How far away from the strips will the wire reach?

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 22

When switch S is closed, current i flows in the wire. The magnetic force acts in rightward due to which rod moves with constant acceleration on rails. When separated from rails, it is retarded because of frictional force acting in opposite direction of motion.

Let it travels × distance before it stops.

Work done by magnetic force= work done by frictional force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 23

A metal wire PQ of mass 10g lies at rest on two horizontal metal rails separated by 4.90cm. A vertically downward magnetic field of magnitude 0.800T exists in the space. The resistance of the circuit is slowly decreased and it is found that when the resistance goes below 20.0Ω, the wire PQ starts sliding on the rails. Find the coefficient of friction.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 23

When wire is about to slide

Frictional force= Magnetic force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 24

A straight wire of length l can slide on two parallel plastic rails kept in a horizontal plane with a separation d. The coefficient of friction between the wire and the rails is µ. If the wire carries a current i, what minimum magnetic field should exist in the space in order to slide the wire on the rails.

Solution 24

When wire is about to slide

Frictional force=magnetic force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 25

Figure shows a circular wire-loop of radius a, carrying a current i, placed in a perpendicular magnetic field B. (a) Consider a small part dl of the wire. Find the force on this part of the wire exerted by the magnetic field. (b) Find the force of compression in the wire.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 25

(a) Force on small segment of wire, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field towards center.

(b) Let a small segment of wire subtends H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field angle at the center of loop.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field angle is small]

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 26

Suppose that the radius of cross-section of the wire used in the previous problem is r. Find the increase in the radius of the loop if the magnetic field is switched off. The Young's modulus of the material of the wire is Y.

Solution 26

In a magnetic field, the wire is in compressed state and when field is removed it will expand as no compression will be there.

We know that,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 27

The magnetic field existing in a region is given by

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

A square loop of edge l and carrying a current i, is placed with its edge parallel to the X-Y axes. Find the magnitude of the net magnetic force experienced by the loop.

Solution 27

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

For wire AB and DC

Magnetic field at distance x is

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Length of element=dx

Current in element=i

Since all parameters are same so magnitude of force is also same but opposite in direction. Hence force on wire AB and DC will cancel out each other.

For wire AB

Magnetic field=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Length of wire=l

Current=i

forceH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (towards left)

For wire BC

Magnetic fieldH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Length of element=l

Current=i

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field (toward right)

Net force on wireH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 28

A conducting wire of length l, lying normal to a magnetic field B, moves a velocity v as shown in figure. (a) Find the average magnetic force on a free electron of the wire. (b) Due to this magnetic force, electrons concentrate at one end resulting in an electric field inside the wire. The redistribution stops when the electric force on the free electrons balances the magnetic force on the free electron balances the magnetic force. Find the electric field developed inside the wire when the redistribution stops. (c) What potential difference is developed between the ends of the wire?

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 28

(a) Force on free electron of wire

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

(b) At equilibrium of electron

Electrostatic force=magnetic force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

(c) Potential differenceH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 29

A current i is passed through a silver strip of width d and width d and area of cross-section A. The number of free electrons per unit volume is n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region as shown in figure, what is the average magnetic force on the free electrons? (c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons. (d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation? The appearance of a transverse emf, when a current-carrying wire is placed in a magnetic field, is called Hall effect.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 29

(a) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

(b) Magnetic force, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

(c) At equilibrium of electron

Electrostatic force = Magnetic force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

(d) potential difference

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 30

A particle having a charge H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldC and a mass of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldg is projected with a speed of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s in a region having a uniform magnetic field of 0.10T. The velocity is perpendicular to the field. Find the radius of the circle formed by the particle and also the time period.

Solution 30

Radius of circle, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Time period, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldsec

Question 31

A proton describes a circle of radius 1cm in a magnetic field of strength 0.10T. What would be the radius of the circle described by an α - particle moving with the same speed in the same magnetic field?

Solution 31

Radius of the circle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Question 32

An electron having a kinetic energy of 100eV circulates in a path of radius 10cm in a magnetic field. Find the magnetic field and the number of revolutions per second made by the electron.

Solution 32

Relation between kinetic energy andradius of circle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT

Number of revolutions per second, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldHz

Question 33

Protons having kinetic energy K emerge from an accelerator as a narrow beam. The beam is bent by a perpendicular magnetic field so that it just misses a plane target kept at a distance l in front of the accelerator. Find the magnetic field.

Solution 33

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Proton just misses the target means

Radius< 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Chapter 12 - Magnetic Field Exercise 233

Question 34

A charged particle is accelerated through a potential difference of 12kV and acquires a speed of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s. It is then injected perpendicularly into a magnetic field of strength 0.2T. Find the radius of the circle described by it.

Solution 34

Work done by potential difference

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Now, radius of the circle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Question 35

Doubly-ionized helium ions are projected with a speed of 10 km/s in a direction perpendicular to a uniform magnetic field of magnitude 0.1T. Find (a) the force acting on an ion, (b) the radius of the circle in which it circulates and (c) the time taken by an ion to complete the circle.

Solution 35

(a) Force=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN

 

(b) Radius of the circle, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

 

(c) Time periodH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldsec

Question 36

A proton is projected with a velocity of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s perpendicular to a uniform magnetic field of 0.6T. find the acceleration of the proton.

Solution 36

Force on the proton

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 37

(a) An electron moves along a circle of radius 1m in a perpendicular magnetic field of strength 0.50T. What would be its speed? Is it reasonable? (b) If a proton moves along a circle of same radius in the same magnetic field, what would be its speed?

Solution 37

(a) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

It is not reasonable as its speed is more than speed of light.

 

(b) For proton

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

Question 38

A particle of mass m and positive charge q, moving with a uniform velocity v, enters a magnetic field B as shown in figure. (a) Find the radius of the circular arc it describes in the magnetic field. (b) Find the angle subtended by the arc at the center. (c) How long does the particle stay inside the magnetic field? (d) Solve the three parts of the above problem if the charge q on the particle is negative.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 38

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

(a) Magnetic force = centripetal force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(b) Angle subtended at center of circle=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(c) Time period H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(d) If charge is negative.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

(i) Radius

Magnetic force = centripetal force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(ii) Angle subtended at center

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(iii) Time period

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 39

A particle of mass m and charge q is projected into a region having a perpendicular magnetic field B. Find the angle of deviation of the particle as it comes out of the magnetic field if the width d of the region is very slightly smaller than

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 39

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

When width of the magnetic field is less than the radius of the circle of particle it will comes and if width of magnetic field is more than radius of circle of particle it will not come out.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(a) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(b) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(c) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

It will not come out from forward direction but will describe a semicircle deviation = π 

Question 40

A narrow-beam of singly-charged carbon ions, moving at a constant velocity of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s, is sent perpendicularly in a rectangular region having uniform magnetic field B=0.5T. It is found that two beams emerge from the field in the backward direction, the separations from the incident beam being 3.0cm and 3.5cm. Identify the isotopes present in the ion beam. Take the mass of an ion=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg, where A is the mass number.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 40

Distance between incident rays and emergent rays=2R

For first beam H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

For second beam H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Radius of the circle is given by

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Mass of 1st particle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldamu

So, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

The two isotopes are of carbon H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field.

Question 41

Fe+ ions are accelerated through a potential difference of 500V and are injected normally into a homogeneous magnetic field B of strength 20.0mT. Find the radius of the circular paths followed by the isotopes with mass number 57 and 58. Take the mass of an ion=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg where A is the mass number.

Solution 41

Work done= chargeH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldpotential difference

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

For isotopes with mass number 57

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

For isotopes with mass number 58

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Question 42

A narrow beam of singly-charged potassium ions of kinetic energy 32keV is injected into a region of width 1.00cm having a magnetic field of strength 0.500T as shown in figure. The ions are collected at a screen 95.5cm away from the field region. If the beam contains isotopes of atomic weights 39 and 41, find the separation between the points where these isotopes strike the screen. Take the mass of a potassium ion=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg where A is the mass number.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 42

Kinetic energy, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

For K-39

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

Time to cross magnetic field

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldsec

Now acceleration in the magnetic field region

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Displacement along y-direction in field

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Velocity in the vertical direction; H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

Time to reach the screen=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldsec

During this time, distance travelled in vertical direction

=speed × time

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Net displacement from horizontal=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Similarly, for K-41

Net displacement from horizontal=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Net gap=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldmm

Question 44

Electrons are emitted with negligible speed from an electron gun are accelerated through a potential difference V along the X-axis. These electrons emerge from a narrow hole into a uniform magnetic field B directed along this axis. However, some of the electrons emerging from the hole make slightly divergent angles as shown in figure. Show that these paraxial electrons are focused on the X-axis at a distance

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 44

Work done= charge × potential difference

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

The electrons which are at slightly divergent angles will follow helical path.

Time period to complete one revolution of helical path

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

In this time period, distance travelled by electron along x-axis will be equal to pitch.

Pitch=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Pitch=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 45

Figure shows a convex lens of focal length 12cm lying in a uniform magnetic field B of magnitude 1.2T parallel to its principal axis. A particle having a charge H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldC and mass H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg is projected perpendicular to the plane of the diagram with a speed of 4.8m/s. The particle moves along a circle with its center on the principal axis at a distance of 18cm from the lens. Show that the image of the particle goes along a circle and find the radius of that circle.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Solution 45

Radius of the circle for object

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Now, magnification

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm negative sign shows that image is inverted.

Chapter 12 - Magnetic Field Exercise 234

Question 45

Two particles, each having mass m are placed at a separation d in a uniform magnetic field B as shown in figure. They have opposite charges of equal magnitude q. At time t=0, the particles are projected towards each other, each with a speed v. Suppose the Coulomb force between the charges is switched off. (a) find the maximum value H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field of the projection speed so that the two particles do not collide. (b) What would be the minimum and maximum separation between the particles if H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field? (c) At what instant will a collision occur between the particles if H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field? (d) Suppose H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field and the collision between the particles is completely inelastic. Describe the motion after the collision.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 45

(a) for particles not to collide

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

d>(R+R)

d>2R

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Maximum speed of particles

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(b)

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Radius of each circle=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Minimum separation between particles=d-2R

=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Maximum separation between particles=d+2R

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(c) Radius od the particle will be

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

The particles will collide after travelling H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field distance in horizontal direction.

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Angle rotated by the particle till collision occurs

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Time taken=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(d)

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

By momentum of conservation, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

along x-axis,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

along y-axis,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

So, particle moves with constant speed along the straight line in upward direction after collision as it becomes chargeless.

Question 46

A uniform magnetic field of magnitude 0.20T exists in space from east to west. With what speed should a particle of mass 0.010g and having a charge H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldC be projected from south to north so that it moves with a uniform velocity?

Solution 46

Particle will move with constant velocity if

Magnetic force=gravitational force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

Question 47

A particle moves in a circle of diameter 1.0cm under the action of a magnetic field of 0.40T. An electric field of 200 V/m makes the path straight. Find the charge/mass ratio of the particle.

Solution 47

Radius of the circle=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

If the particle moves with constant velocity

Electrostatic force= magnetic force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldC/Kg 

Question 48

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field m/s. The velocity is perpendicular to both the fields. When the electric field is switched off, the proton moves along a circle of radius 4.0cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of proton H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg.

Solution 48

Initially,

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

When electric field is switched off, the proton will move in a circle due to force by magnetic field.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

magnetic field, B=0.05T

electric field, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN/C

Question 49

A particle having a charge of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field and a mass of H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg is projected with a speed of 1.0km/s in a magnetic field of magnitude 5.0mT. The angle between the magnetic field and the velocity is H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field. Show that the path of the particle will be a helix. Find the diameter of the helix and its pitch.

Solution 49

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldC

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldT

Component of velocity along magnetic field=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Component of velocity perpendicular to magnetic field=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Radius of the particle, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Diameter=2R=2(0.18) =0.36m=36cm

Pitch=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldcm

Question 50

A proton projected in a magnetic field of 0.020T travels along a helical path of radius 5.0cm and pitch 20cm. Find the components of the velocity of the proton along and perpendicular to the magnetic field. Take the mass of the proton=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg.

Solution 50

m=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldkg

B=0.02T

R=5cm=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

P=20cm=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm

Let velocity of the particle perpendicular and parallel to magnetic field be H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field respectively.

Radius of helical path

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

Pitch of helical path

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldm/s

Question 51

A particle having a mass m and charge q is released from the origin in a region in which electric field and magnetic field are given by H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field and H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field . Find the speed of the particle as a function of its z-coordinate.

Solution 51

Let velocity of the particle after it travelled distance 'z' along z-axis.

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Force on the particle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Since particle is in y-z plane only, So H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Force along z-axis

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Also, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 52

An electron is emitted with negligible speed from the speed from the negative plate of a parallel plate capacitor charged to a potential difference V. The separation between the plates is d and a magnetic field B exists in the space as shown in figure. Show that the electron will fail to strike upper plate if

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

 

Solution 52

Electric field setup between the plates of a capacitor, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Work done by electric field = change in kinetic energy

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

The electron will move in circular path of radius, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Since electron doesn't strike upper plate

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 53

A rectangular coil of 100 turns has length 5cm and width 4cm. It is placed with its plane parallel to a uniform magnetic field and a current of 2A is sent through the coil. Find the magnitude of the magnetic field B, if the torque acting on the coil is 0.2N-m.

Solution 53

N=100 turns

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

I=2A

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN-m

We know that, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

B=0.5T

Question 54

A 50-turn circular coil of radius 2.0cm carrying a current of 5.0A is rotated in a magnetic field of strength 0.20T. (a) What is the maximum torque that acts on the coil? (b) In a particular position of the coil, the torque acting on it is half of this maximum. What is the angle between the magnetic field and the plane of the coil?

Solution 54

(a) H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN-m

(b) Let angle between magnetic field and plane of coil be H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 55

A rectangular loop of sides 20cm and 10cm carries a current of 5.0A. A uniform magnetic field of magnitude 0.20T exists in parallel to the longer side of the loop. (a) What is the force acting on the loop? (b) What is the torque acting on the loop?

Solution 55

(a) Force acting on both longer sides is zero as angle between length of wire and magnetic field is zero.

Force on shorter sides are equal in magnitude but opposite in direction. Hence will cancel out each other.

So, net force on loop is zero.

(b) torque, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN-m

Question 56

A circular coil of radius 2.0cm has 500 turns in it and carries a current of 1.0A. Its axis makes an angle of 30° with the uniform magnetic field of magnitude 0.40T that exists in the space. Find the torque acting on the coil.

Solution 56

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

N=500 turns

I=1A

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Torque, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic FieldN

Question 57

A circular loop carrying a current i has wire of total length L. A uniform magnetic field B exists parallel to the plane of the loop. (a) Find the torque on the loop. (b) If the same length of the wire is used to form a square loop, what would be the torque? Which is larger?

Solution 57

(a) Length of wire=circumference of circle

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Area of circular loopH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Torque, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(b) Length of the wire= perimeter of square of side 'a'

L=4a

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Area of square=H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Torque, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Torque on circular loop is larger than square loop.

Question 58

A square coil of edge l having n turns carries a current i. It is kept on a smooth horizontal plate. A uniform magnetic field B exists in a directional parallel to an edge. The total mass of the coil is M. What should be the minimum value of B for which the coil will start tipping over?

Solution 58

Torque by the magnetic field

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

For the coil to start tipping over

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldtorque due to gravitational force

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Minimum magnetic field H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 59

Consider a nonconducting ring of radius r and mass m which has a total charge q distributed uniformly on it. The ring is rotated about its axis with an angular speedω. (a) Find the equivalent electric current in the ring, (b) find the magnetic moment H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Fieldof the ring. (c) Show that H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field where l is the angular momentum of the ring about its axis of rotation.

Solution 59

(a) Current, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(b) magnetic moment, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

(c) angular momentum

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Putting this value in magnetic moment expression

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Chapter 12 - Magnetic Field Exercise 235

Question 60

Consider a nonconducting plate of radius r and mass m which has a charge q distributed uniformly over it. The plate is rotated about its axis with an angular speedω. Show that the magnetic moment µ and the angular momentum l of the plate are related as H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field.

Solution 60

Consider a ring at distance x from center and of width dx

 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Charge on the elemental ring H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Current due to rotation of this elemental ring

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Magnetic moment due to elemental ring H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Total magnetic momentH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Angular momentum

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Divide (1) and (2)

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Question 61

Consider a solid sphere of radius r and mass m which has a charge q distributed uniformly over its volume. The sphere is rotated about a diameter with an angular speed ω. Show that the magnetic moment H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field and the angular momentum l of the sphere are related as H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field.

Solution 61

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field

 

Consider a disc at distance x from center of sphere of thickness 'dx'

Consider an elemental ring at distance y of thickness 'dy' from center of disc

Charge on elemental ringH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Current due to elemental ringH-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Magnetic moment due to elemental ring

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Magnetic moment for complete sphere

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

Angular momentum, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

So, H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field 

H-c-verma Solutions Cbse Class 12-science Physics Chapter - Magnetic Field