Class 12-science H C VERMA Solutions Physics Chapter 13 - Magnetic Field due to Current
Magnetic Field due to Current Exercise 249
Solution 1
Unit of B 
Now, for 
Unit of 
Solution 2
Magnetic field due to wire
T
Field is along positive y-axis by the right hand thumb rule.
Magnetic Field due to Current Exercise 250
Solution 3
Maximum magnitude of magnetic field will be at the surface of wire i.e., at distance r=0.8mm from center of wire.
mT
Solution 4
Magnetic field due to wire
mT
Solution 5
Magnetic field due to the wire is given by
T
Net magnetic field at point P
T
Net magnetic field at point Q
Solution 6
(a)
Magnetic field due to wire is maximum at the surface of wire.
(b) magnetic field due to wire is maximum at the surface
If 
So, If 
then minimum magnetic field 
If 
then minimum magnetic field is 
Solution 7
Magnetic field due to wire
T
Now, field due to wire and external field are perpendicular to each other.
T
Solution 8
Let at distance r from wire in the west direction net magnetic field is zero.
So, 
mm
Solution 9
Magnetic field due to long wire
At point 
T
T
T
At point 
T
T
T
At point 
T
T
T
At point 
T
Both fields are perpendicular to each other
T
Solution 10
Magnetic field due to wire P and Q
T
T
Solution 11
(a)
Magnetic field due to wires are of equal magnitude but in opposite direction.
So, resultant field=0.
(b)
Magnetic field due to both wires is in upward direction and is of equal magnitude.
T
(c)
Magnetic field due to wires are of equal magnitude but in opposite direction.
So, resultant field=0.
(d)
Magnetic field due to both wires is in upward direction and is of equal magnitude.
T
Solution 12
Magnetic field due to 
-long wire=
Let magnetic field due
to wire (1), (2), (3) and (4) be 
respectively.
The magnetic field due all four wires at point P is equal in magnitude.
The direction of
magnetic fields are
.
So, net magnetic field=0
At point 
At point 
But opposite in direction, so will cancel
each other.
But opposite in direction, so will cancel
each other.
So, net magnetic field=0
At point 
Solution 13
Magnetic field at the axial point of wire is zero.
Magnetic field due to
semi-infinite wire is=
Magnetic field due to
wire (1) and wire (2) be 
and 
respectively.
At point P
(upward)
(upward)
At point Q
(downward)
(downward)
At point R
(downward)
(downward)
At point S
(upward)
(upward)
Solution 14
Here, 
Magnetic field at point P
When d>>>x (neglecting x)
When d<<<x (neglecting d)
Solution 15
Magnetic field at P
T
Solution 16
Magnetic field due to
infinite wire 
Magnetic field due to
section 
Here,
On solving,
Solution 17
Wire ABC and wire ADC are in parallel combination.
So, current will be distributed in inverse ratio of resistance.
Magnetic field due to AB and BC at center of loop
Net field at center=
Solution 18
The current will be equally distributed in wire ABC and ADC as both are in parallel combination and have same resistance.
Now,
But in opposite direction hence will cancel
each other.
(upward direction)
(downward direction)
Net magnetic field
Magnetic Field due to Current Exercise 251
Solution 19
The current in wire ADCB and wire AB will be in inverse ratio of resistance.
Magnetic field at center of loop
Solution 20
The current is equally divided in both sides.
But in opposite direction.
But in opposite direction.
So, net magnetic field=0
Solution 21
(a)
Perpendicular distance
of point O from wire PR=
Net magnetic field, 
(b)
Solution 22
Magnetic field at P due to single wire
Net magnetic field 
Solution 23
Solution 24
Length of each side, 
Angle by each side at
center 
Magnetic field by a side at center
Magnetic field by all sides
Solution 25
Net current in the circuit=0
So, magnetic field is zero at point P.
Solution 26
Force per unit length between parallel wires
cm
Solution 27
On wire (1)
N towards middle wire
On wire (2)
But opposite in direction.
On wire (3)
N towards middle wire
Solution 28
Let third wire is placed in middle of both wires at a distance of x from wire carrying current of 10A
Since, no magnetic force
x=2cm
Solution 29
Current in each wire will be same as all have same resistance.
So, current= 10A in each wire.
For wire AB
Force due to CD wire +
Force due to EF wire.
For wire CD,
Force due to AB is equal to force due to FF wire but in opposite direction.
Solution 30
Magnetic force between wires=weight of second wire
A
Solution 31
(a)At distance x from wire consider two small elements of length dx in wire PQ and SR.
Magnetic field is same at both location.
Force on element in PQ is in upward direction and force on element in SR is in downward direction but in opposite direction.
So,
(b) magnetic field at SP
by wire
T
Force of SP wire=
N (in right direction)
magnetic field at RQ by
wire
T
Force of SP wire=
N (in left direction)
So, net force
N
Solution 32
Magnetic field for a circular loop
cm
Solution 33
Magnetic field for a circular loop
mA
Solution 34
Magnetic field at center
T
Solution 35
Current in each
semi-circle=
Magneti field due to each semicircle is same but in opposite direction. So, net magnetic field is zero.
Magnetic Field due to Current Exercise 252
Solution 36
Magnetic field due to smaller coil
T
Magnetic field due to bigger coil
T
(a) for same direction
T
(b) for opposite direction
Solution 37
Now, magnetic field due to both coils will be in perpendicular direction.
mT
Solution 38
r=20cm=0.2m
N
Solution 39
Magnetic field due to
larger loop at center=
Torque acting on smaller loop
Solution 40
Magnetic field due to
larger loop at center=
Torque acting on smaller loop
Since, smaller loop is at rest
Solution 41
Magnetic field due to semi-circle
T
Solution 42
T
Solution 43
Let straight wire is placed at distance x from center of loop.
Since, magnetic field at center is zero.
Solution 44
N=200 turns; R=0.1m;I=2A
(a) 
mT
(b) 
Here, 
x=7.66cm
Solution 45
(a) magnetic field along axis of a circular loop is given by
T (downward direction)
(b) magnetic field along axis of a circular loop is given by
T (downward direction)
Solution 46
C
r=20cm=0.2m
sec
A
Electric field, 
Magnetic field,
m/s
Solution 47
(a) The current enclosed by ampere loop of radius r/2 is zero.
So, magnetic field is also zero.
(b) Magnetic field outside hollow cylindrical tube
Solution 48
(a) The current enclosed by ampere loop inside the tube is zero. So, magnetic field is also zero.
(b) Consider an ampere loop just outside the tube of radius b.
Solution 49
Consider an ampere loop of radius 'a' with center on the axis of cylinder.
Current enclosed=
By ampere's circuital law
Solution 50
(a) Magnetic field inside solid wire of radius 'b' at a distance of 'a' from center of the axis of cylinder.
T
(b) magnetic field at the surface of wire
T
(c) magnetic field at the outside of wire
T
Solution 51
Consider a point A inside the loop PQRS where B=0.
According to Ampere's circuital law,
If there is current enclosed by the loop PQRS, then magnetic field B cannot be zero.
Whereas, we have taken magnetic field at point A to be zero.
Thus, such field is not possible.
Solution 52
Magnetic field due to a sheet at a distance 'd' from it having current through a strip of width dl is 'Kdl' can be calculated by ampere's circuital law.
(independent of distance)
At point P and R
Field due to both sheets are equal in magnitude but opposite in direction. Hence, net field is zero.
At point Q
Field due to both sheets are equal and is in same direction.
Solution 53
We know that,
Solution 54
Magnetic field for solenoid
turns/m
Solution 55
Width of each turn=1mm=
m
No. of turns=
T
Solution 56
T
No. of turns per unit
length 
Resistance of the wire
Emf
V=IR
volt
Magnetic Field due to Current Exercise 253
Solution 57
(a) Magnetic field at the axis due to the circular loop
Here, radius=a distance
of the center of the solenoid from the center of circular current=
For the whole solenoid
(b) when l>>>a
So,
If a>>>l
So, 
Solution 58
Frequency of particle in uniform magnetic field
Solution 59
The particle will undergo uniform circular motion. For particle not to strike with solenoid, radius is r/2.
Radius=
Solution 60
Magnetic field due to
the metal sheet 
Magnetic field due to
solenoid 
(a) Magnetic field is zero.
So, 
(b) Now, field due to sheet and solenoid is perpendicular
Solution 61
C=100
; V=20volt
Q=CV
C
Now, potential drops to 90%
volt
C
Average current flown=
A
Magnetic field at center of solenoid
T
