10 Magnetic Fields due to Electric Current
Multiple Choice Questions
1. Speed of an electron passing undeviated through a region of cross electric and magnetic field of magnitude 
and 
respectively in metre per second is(A)
(B)
(C)
(D)
Ans. (C) 
2. If ‘ 
‘ is the radius of dees and ‘ 
‘ be the magnetic field of induction in which positive charge (q) of mass (m) escapes from the cyclotron, then its maximum speed 
is
(A) 
(B) 
(C) 
(D) 
Ans. (C) 
3. A solenoid of 
length and 
in diameter possess 10 turns per metre. A current of 
is flowing through it. The magnetic induction at a point inside the solenoid along the axis is 
(A) 
(B) 
(C) 
(D) 
Ans. (B) 
4. Which one of the following particles cannot be accelerated by a cyclotron?
(A) Electrons
(B) Protons
(C) Deuterons
(D) 
-particles
Ans. (A) Electrons
5. Cyclotron can not accelerate
(A) protons
(B) neutrons
(C) 
-particles
(D) deuterons
Ans. (B) neutrons
6. The unit 
is equal to
(A) henry
(B) watt
(C) dyne
(D) tesla
Ans. (D) tesla
7. The magnitude of the magnetic field at the centre of a circular current carrying coil varies
(A) inversely with the square of the radius of the coil
(B) directly with the radius of the coil
(C) inversely with the radius of the coil
(D) directly with the square of the radius of the coil
Ans. (C) inversely with the radius of the coil
Theory Questions
10.6 Force on a Closed Circuit in a Magnetic Field 
- What is the value of force on a closed circuit in a magnetic field?
Ans: Force on a closed circuit in a magnetic field is zero.
10.7 Torque on a Current Loop
- State the principle of a moving coil galvanometer and explain its working.
Ans: Principle of M.C.G:
When a coil carrying an electric current is suspended in a uniform magnetic field, a torque acts on it. This torque tends to rotate the coil about the axis of suspension so that the magnetic flux passing through the coil is maximum.
Working:
i. M.C.G. consists of a coil of several turns mounted (suspended or pivoted) in such a way that it can freely rotate about a fixed axis, in a radial uniform magnetic field.
ii. A soft iron cylindrical core makes the field radial and strong.
iii. The coil rotates due to a torque acting on it as the current flows through it. Torque acting on current carrying coil is 
NIAB 
.
Here 
as the field is radial.
where A is the area of the coil, B the strength of the magnetic field, 
the number of turns of the coil and I the current in the coil.
iv. This torque is counter balanced by a torque due to a spring fitted at the bottom so that a fixed steady current 
in the coil produces a steady angular deflection 
.
v. Larger the current is, larger is the deflection and larger is the torque due to the spring. If the deflection is 
, the restoring torque due to the spring is equal to 
where 
is the torsional constant of the spring.
Thus, 
,
and the deflection 
This means, the deflection 
is proportional to the current I i.e., 
I.
2 Draw a neat and labelled diagram of suspended coil type moving coil galvanometer.
Ans. Refer Subtopic 10.7: Q. No. 1 (Diagram only)
10. 15 Ampere’s Law
1 State and explain Ampere’s circuital law.
Ans. : Statement:
The line integral of magnetic field of induction 
around amy closed path in free space is equal to boohe permeability of free space 
times the total arrent flowing through area bounded by the path. Monthematically, 
The sign 
indicates that the integral is to be evaluated over a closed loop called Amperian loop. The current I on the right hand side is the net current encircled by the Amperian loop.
Explanation:
i. Consider cross-sections of four long straight wires carrying currents 
into or out of the plane of the paper are shown in figure below.
ii. An Amperian loop is drawn to encircle 3 of the current wires and not the fourth one.
iii. As the current goes perpendicular to the plane of the paper, 
is in the plane of the paper even if its direction is unknown.
iv. The length element on the Amperian loop is 
(in the plane of the paper).
,
Using Ampere’s law, 
v. Thus the integration is over the product of length 
of the Amperian loop and the component of the magnetic field 
, tangent to the loop.
vi. Using curled palm right hand thumb rule, for the distribution of currents as shown 
and 
are coming out of the paper, 
parallel to the stretched thumb. Hence these are positive. 
, on the other hand is going into the plane of the paper 
. Thus, it is negative.
The Current 
is not within the Amperian loop. As the integration is over a full loop, contributions of 
to 
cancel out.
- State Ampere’s circuital law.
Using Ampere’s circuital law, obtain an expression for magnetic induction at any point due to a straight conductor carrying current.
OR
State Ampere’s law and hence obtain an expression for the magnetic induction at any point near a straight conductor carrying a current.
Ans: Refer Subtopic 10.15: Q. No. 1 (statement only) Expression for magnetic induction at any point due to a straight conductor carrying current:
i. Consider a long straight wire carrying a current I as shown in figure below.
and 
are tangential to the Amperian loop which in this case is a circle.
ii. The field 
at a distance 
from the wire is given by B 
This is the required expression.
10.16 Magnetic Field of a Solenoid and a Toroid
- State Ampere’s circuital law. Obtain an expression for magnetic induction along the axis of toroid.
Ans: Refer Subtopic 10..15: Q. No. 1 (Statement only) Expression for magnetic induction at a point along the axis of a toroid:
i. Consider a toroid carrying steady current I and having turns 
.
ii. The magnetic field around the toroid consists of concentric circular lines of force around it. Magnetic field is produced, when a steady current ‘ 
‘ flows through toroid.
iii. The direction of magnetic field at a point is along the tangent to the circular path at that point.
iv. Let 
be the radius of the Amperian loop. This loop is concentric with the axis of toroid. 
is a point on the loop.
v. Applying Ampere’s law,
where,
total number of turns in the toroid.
total current flowing through toroid.
Current flowing in each turn
Now, 
vi. But, 
and 
are in same direction
Equation (2) can be written as,
vii. Also 
….(total length of circle is its circumference)
From equation (3),
viii. From equation (1) and (4),
B 
This is the required expression.
- Obtain an expression for magnetic induction of a toroid of ‘
‘ turns about an axis passing through its centre and perpendicular to its plane.
‘ turns about an axis passing through its centre and perpendicular to its plane.OR
Obtain an expression for magnetic induction along the axis of toroid on the basis of Ampere’s circuital law.
Ans: Refer Subtopic 10..16: Q. No. 1 for expression for magnetic induction along the axis of toroid on the basis of Ampere’s circuital law.
From formula (ii),
Ans: The radius of the circular path in cyclotron is 
and the frequency needed for the applied alternating voltage of the parallel resonant cyclotron is 
.
10.7 Torque on a Current Loop
- A closely wound solenoid of 1000 turns and area of cross-section
carries a current of 
. It is placed in a horizontal plane with its axis making an angle of 
with the direction of uniform magnetic field of 
. Calculate the torque acting on the solenoid.
carries a current of 
. It is placed in a horizontal plane with its axis making an angle of 
with the direction of uniform magnetic field of 
. Calculate the torque acting on the solenoid.Solution:
Given:
,
Since the axis of the coil is kept inclined at 
with the direction of uniform magnetic field, 
To find: 
Torque acting on solenoid 
Formula: 
Calculation: From formula,
Ans: The torque acting on the solenoid is 
.
- A coil has ‘
‘ turns, each of cross-sectional area 
. The axis of the coil is kept inclined at with the direction of uniform magnetic field of induction 
. The torque of 
is experienced by the coil, when a current of 1.25 A flows through each turn. Calculate n.
‘ turns, each of cross-sectional area 
. The axis of the coil is kept inclined at 
with the direction of uniform magnetic field of induction 
. The torque of 
is experienced by the coil, when a current of 1.25 A flows through each turn. Calculate n. Solution:
Given:
Since the axis of the coil is kept inclined at 
with the direction of uniform magnetic field, 
To find: . Number of turns (n)
Formula: 
Calculation: From formula,
Using reciprocal table.
Ans: Number of turns in coil are around 58.
- A moving coil galvanometer (M.C.G.) has 10 turns each of length
and breadth 
. The coil of M.C.G. carries a current of 
and is kept perpendicular to the uniform magnetic field of induction 
. The twist constant of phosphor bronze fibre is 
degree. Calculate the deflection produced.
and breadth 
. The coil of M.C.G. carries a current of 
and is kept perpendicular to the uniform magnetic field of induction 
. The twist constant of phosphor bronze fibre is 
degree. Calculate the deflection produced.Solution:
Given: 
,
degree
To find: 
Deflection 
Formula: 
Calculation: From formula,
Ans: The deflection produced is 
.
10.8 Magnetic Dipole, Moment
- A circular coil of 250 turns and diameter carries a current of
. What is the magnitude of magnetic moment associated with the coil?
carries a current of 
. What is the magnitude of magnetic moment associated with the coil?Solution:
Given: 
To find: 
Magnetic moment of the coil (m)
Formula: 
NIA
Calculation: For the coil,
Using formula,
antilog 
antilog 
antilog 
Ans: Magnetic moment of the coil is 
.
- A circular coil of 300 turns and average area carries a current of . Calculate the magnitude of magnetic moment associated with the coil.
carries a current of 
. Calculate the magnitude of magnetic moment associated with the coil.Solution:
Given: 
To find: 
Magnetic moment of the coil (m)
Formula: ‘ 
NIA
Calculation: From formula,
Ans: The magnitude of magnetic moment associated with coil is 
.
10.10 Magnetic Field due to a Current: Biot-Savart Law
- Calculate the value of magnetic field at a distance from a very long, straight wire carrying a current of .
from a very long, straight wire carrying a current of 
.Solution:
Given:
To find:
Magnitude field (B)
Formula: 
Calculation: From formula,
Ans: The value of magnetic field is 
.
10.11 Force of Attraction between two Long Parallel Wires
- A current of equal magnitude flows through two long parallel wires separated by . If force per unit length of
acts on both the wires respectively, calculate the current through each wire.
. If force per unit length of 
acts on both the wires respectively, calculate the current through each wire.Solution:
Given:
To find:
Electric current (I)
Formula: 
Calculation: From formula,
Ans: The electric current through each wire is 
.
- Calculate the current flowing through two long parallel wires carrying equal currents and separated by a distance of
experiencing a force per unit length of 
.
experiencing a force per unit length of 
.Solution:
Given:
To find:
Formula: 
Calculation: From formula,
Ans: The electric current is 
.
10.16 Magnetic Field of a Solenoid and a Toroid
- A solenoid
long and 
in diameter has two layers of windings of 500 turns each and carries a current of . Calculate the magnetic induction at its centre along the axis. [Mar 15]
long and 
in diameter has two layers of windings of 500 turns each and carries a current of 
. Calculate the magnetic induction at its centre along the axis. [Mar 15]Solution:
Given: 
2 layers of windings, hence total turns 
To find: 
Magnetic induction (B)
Formula: 
Calculation: From formula,
Ans: The magnetic induction at centre of solenoid along the axis is 
.