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

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231

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Magnetic Field Exercise 230

Solution 1

N towards west.

Solution 2

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

(b) Kinetic energy,

(horizontally)

Magnetic force,

(along west direction)

Time taken to travel 1m in horizontal direction

In this time, deflection of electron in west direction

deflection

deflection=

deflectionm

deflectioncm

Solution 3

Let velocity of the particle in X-Y plane

Force on particle,

Comparing,

Velocity,

Solution 4

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

So,

So, acceleration of particle is .

Solution 5

Time to travel 100m distance horizontally,

sec

Force on the bullet

Deflection due to the magnetic field

deflection

deflection=

deflectionm

Solution 6

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

So, initially force have been acted by electric 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).

(into the paper perpendicularly)

Solution 7

Force=i(l × B)

F ≈ 0.08N actin perpendicular to both wire and 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,

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.

Magnetic Field Exercise 231

Solution 9

Length of the wire in magnetic field=m

Force of wire,

Solution 10

Solution 11

Force,

N upward in the plane of paper.

Solution 12

Magnetic field is radially outwards.

Magnetic field and length are perpendicular to each other.

Now,

Solution 13

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

Solution 14

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.

So, initially Tension

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

So, finally Tension

Change in Tension

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 is

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.

Solution 16

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

Considering two element at an angle as shown in figure.

Force acting on each element

Resultant force on both element

Resultant force

(distance between initial and final point)

Solution 17

Length of wire

Magnetic field, B=0.5T

Current, i=5A

Force

F=0.25N

Solution 18

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

Length of wire

Magnetic force,

Solution 19

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

Length of wire=2R

Force=

F=2iRB

Solution 20

Under equilibrium

Magnetic force=Gravitational force

Solution 21

(a) Switch 'S' is open

Weight of the rod is balanced by the tension in threads

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

T=1.25N

Magnetic Field Exercise 232

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

Solution 23

When wire is about to slide

Frictional force= Magnetic force

Solution 24

When wire is about to slide

Frictional force=magnetic force

Solution 25

(a) Force on small segment of wire,

towards center.

(b) Let a small segment of wire subtends angle at the center of loop.

angle is small]

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,

Solution 27

For wire AB and DC

Magnetic field at distance x is

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=

Length of wire=l

Current=i

force

(towards left)

For wire BC

Magnetic field

Length of element=l

Current=i

(toward right)

Net force on wire

Solution 28

(a) Force on free electron of wire

(b) At equilibrium of electron

Electrostatic force=magnetic force

Solution 29

(a)

(b) Magnetic force,

Electrostatic force = Magnetic force

(d) potential difference

Solution 30

Radius of circle,

Time period,

sec

Solution 31

Radius of the circle

Solution 32

Relation between kinetic energy andradius of circle

Number of revolutions per second,

Solution 33

Proton just misses the target means

Radius<

Magnetic Field Exercise 233

Solution 34

Work done by potential difference

Now, radius of the circle

Solution 35

(a) Force=

(b) Radius of the circle,

sec

Solution 36

Force on the proton

Solution 37

(a)

m/s

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

(b) For proton

m/s

Solution 38

(a) Magnetic force = centripetal force

(b) Angle subtended at center of circle=

(d) If charge is negative.

(i) Radius

Magnetic force = centripetal force

(ii) Angle subtended at center

(iii) Time period

Solution 39

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.

(a)

(b)

(c)

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

Solution 40

Distance between incident rays and emergent rays=2R

For first beam cm

For second beam cm

Radius of the circle is given by

Mass of 1^st particle

amu

So,

The two isotopes are of carbon .

Solution 41

Work done= chargepotential difference

For isotopes with mass number 57

For isotopes with mass number 58

Solution 42

Kinetic energy,

For K-39

m/s

Time to cross magnetic field

sec

Now acceleration in the magnetic field region

Displacement along y-direction in field

Velocity in the vertical direction;

m/s

Time to reach the screen=sec

During this time, distance travelled in vertical direction

=speed × time

Net displacement from horizontal=

Similarly, for K-41

Net displacement from horizontal=m

Net gap=mm

Solution 43

Radius of the circle for object

Now, magnification

cm negative sign shows that image is inverted.

Solution 44

Work done= charge × potential difference

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

Time period to complete one revolution of helical path

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

Pitch=

Magnetic Field Exercise 234

Solution 45

(a) for particles not to collide

d>(R+R)

d>2R

Maximum speed of particles

(b)

Radius of each circle=

Minimum separation between particles=d-2R

Maximum separation between particles=d+2R

The particles will collide after travelling distance in horizontal direction.

Angle rotated by the particle till collision occurs

Time taken=

(d)

By momentum of conservation,

along x-axis,

along y-axis,

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

Solution 46

Particle will move with constant velocity if

Magnetic force=gravitational force

m/s

Solution 47

Radius of the circle=cm

If the particle moves with constant velocity

Electrostatic force= magnetic force

C/Kg

Solution 48

Initially,

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

magnetic field, B=0.05T

electric field, N/C

Solution 49

m/s

Component of velocity along magnetic field=

Component of velocity perpendicular to magnetic field=

Radius of the particle,

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

Pitch=

Solution 50

m=kg

B=0.02T

R=5cm=m

P=20cm=m

Let velocity of the particle perpendicular and parallel to magnetic field be respectively.

Radius of helical path

m/s

Pitch of helical path

m/s

Solution 51

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

Force on the particle

Since particle is in y-z plane only, So

Force along z-axis

Also,

Solution 52

Electric field setup between the plates of a capacitor,

Work done by electric field = change in kinetic energy

The electron will move in circular path of radius,

Since electron doesn't strike upper plate

Solution 53

N=100 turns

I=2A

N-m

We know that,

B=0.5T

Solution 54

(a)

N-m

(b) Let angle between magnetic field and plane of coil be

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,

N-m

Solution 56

N=500 turns

I=1A

Torque,

Solution 57

(a) Length of wire=circumference of circle

Area of circular loop

Torque,

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

L=4a

Area of square=

Torque,

Torque on circular loop is larger than square loop.

Solution 58

Torque by the magnetic field

For the coil to start tipping over

torque due to gravitational force

Minimum magnetic field

Solution 59

(a) Current,

(b) magnetic moment,

Putting this value in magnetic moment expression

Magnetic Field Exercise 235

Solution 60

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

Charge on the elemental ring

Current due to rotation of this elemental ring

Magnetic moment due to elemental ring

Total magnetic moment

Angular momentum

Divide (1) and (2)

Solution 61

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 ring

Current due to elemental ring

Magnetic moment due to elemental ring

Magnetic moment for complete sphere

Angular momentum,

So,

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

Magnetic Field Exercise 230

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Magnetic Field Exercise 231

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Magnetic Field Exercise 232

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Magnetic Field Exercise 233

Solution 34

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Solution 43

Solution 44

Magnetic Field Exercise 234

Solution 45

Solution 46

Solution 47

Solution 48

Solution 49

Solution 50

Solution 51

Solution 52

Solution 53

Solution 54

Solution 55

Solution 56

Solution 57

Solution 58

Solution 59

Magnetic Field Exercise 235

Solution 60

Solution 61

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