CHAPTER FOUR

MOVING CHARGES AND MAGNETISM

Introduction
1n 1820, a Danish physicist Hans Christian Oersted observed that a magnetic
field is associated with electric current. He showed that electric current
passing through a straight wire can produced deflection in the magnetic
compass needle kept close to it.
In oersted experiment a straight wire carrying current is kept parallel to a
pivoted magnetic needle.

It is observed that the magnetic needle gets deflected. When the direction of
the current is reversed the magnetic needle gets deflected in the opposite
direction.
The direction of deflection of the needle can be determined using Ampere
swimming rule.
Imagine a man swimming along the wire in the direction of the current with his
face always turned towards the needle, then north pole of the needle will be
deflected towards his left hand.
It was also observed that when iron filing sprinkled around the wire it arranges
themselves into concentric circles with the wire as centre in the plane
perpendicular to the wire.

This shows that the pattern of magnetic field lines are concentric circles and
the direction of magnetic field is determined by right hand thumb rule.
Magnetic field
Just as a static charge produces electric field, a moving charge or current
through a conductor produces electric field as well as magnetic field.

1
The magnetic field is the space around a conductor carrying current or the
space around a magnet in which its magnetic effect can be felt.
⃗.
Magnetic field is a vector quantity and is represented by 𝐵
Force or a moving charge in a magnetic field.
The electric charge moving in a magnetic field experiences a force. This fact
was first recognized by Dutch physicist Hendrick Antoon Lorentz

⃗ with a
Consider a positive charge q moving in a uniform magnetic field 𝐵
velocity 𝑣. Let the angle between 𝑣 and 𝐵⃗ be 𝜃.

Due to the interaction between the magnetic field produced by the moving
charge and magnetic field applied, the charges q experiences a force, which
depends on the following.
The magnitude of the force is
(i) directly proportional to charge q
F∝q
(ii) directly proportional to the magnetic field⃗⃗⃗𝐵 .
F∝𝐵 ⃗
(iii) directly proportional to the component of velocity perpendicular to the
direction of magnetic field
F ∝ vsin 𝜃
Combining
F ∝ q𝐵 ⃗ vsin 𝜃
F = q (𝑣 × 𝐵 ⃗)
Thus the direction of the force is perpendicular to both v and magnetic field
and is given by flemings left hand rule.
SI unit of magnetic field
We have

2
𝐹 = q (𝑣 × ⃗⃗⃗𝐵 ) = q𝐵
⃗ v sin 𝜃
𝐹
⃗ =
𝐵
q vsin 𝜃
⃗ =𝐹
If q = 1, v = 1 and 𝜃 = 90 then 𝐵
Thus, magnetic field at a point is defined as the force acting on a unit charge
moving with unit velocity at right angles to the direction of the field.
𝐹
⃗ =
𝐵
q vsin 𝜃
𝑁
⃗ =
𝐵 = NA-1m-1= Tesla (T)
C m 𝑠 −1 sin90
One Gauss = 10-4 T
Lorentz force
The total force experienced by a charged particle moving in a region where
both electric and magnetic fields are present is known as Lorentz force
A charge q in an electric field experiences an electric force
⃗⃗⃗
𝐹𝑒 = q 𝐸⃗
The magnetic force experienced by the charge q moving with velocity v in the
magnetic field
⃗⃗⃗⃗
𝐹𝑚 = q (𝑣 × 𝐵⃗)
Lorentz force F = q 𝐸⃗ + q (𝑣 × 𝐵 ⃗)
Show that no work is done by this force on the charged particle.
Magnetic Lorentz force
⃗⃗⃗⃗
𝐹𝑚 = q (𝑣 × 𝐵⃗)
Since force is perpendicular to velocity
Force is perpendicular to displacement made by the charged particle.
Work done W = F d cos 𝜃= F d cos 90 = 0
So no work is done by magnetic lorentz force on charged particle.
Motion of a charged particle in a uniform magnetic field.
Consider a charged particle having a charge q movoing with a velocity 𝑣 enter a
magnetic field 𝐵⃗ . It experience a magnetic force
F = q ⃗⃗⃗⃗ ⃗ )= q v B sin 𝜃
(𝑣 × 𝐵
Case 1
When the initial velocity is parallel or anti parallel to the magnetic field
𝜃 = 0 𝑜𝑟 180

3
F = q ⃗⃗⃗⃗ ⃗ )= q v B sin 𝜃= 0
(𝑣 × 𝐵
Here the magnetic field does not exert any force on the charged particle. The
charged particle will continue to move along the line of force.
Case 2
When the initial velocity is perpendicular to the magnetic field.
𝜃 = 90
F = q ⃗⃗⃗⃗ ⃗ )= q v B sin 𝜃= q v B sin 90 = q v B
(𝑣 × 𝐵
This force is always perpendicular to the direction of motion and the magnetic
field. So this force continously deflects the particle side ways without changing
its speed and the charged particle will move along a circle perpendicular to the
field.

Here the force F provides the required centripetal force for motion along the
circular path of radius r.
𝑚𝑣 2
qvB= 𝑟
𝑚𝑣 2 𝑚𝑣
r= =
𝑞𝑣𝐵 𝑞𝐵
2𝜋𝑟 2𝜋𝑚𝑣 2𝜋𝑚
The time period T = = =
𝑣 𝑞𝑣𝐵 𝑞𝐵
The time period is independent of v and r.
If the particle moves faster the radius becomes larger so that the time taken
will remain the same.
Angular frequency,
2𝜋 2𝜋𝑞𝐵 𝑞𝐵
𝜔 = 2𝜋𝑓 = = =
𝑇 2𝜋𝑚 𝑚
Which is also independent of the velocity. This fact has important application
in the design of particle accelerator cyclotron.
𝑞
All the charged particles having the same specific charge 𝑚 but moving with
different velocities at a point will complete their circular paths in the same
time.
Case 3
When the intial velocity makes an arbitrary angle with the direction of the
magnetic field.

4
Suppose a charged particle of mass m and charge q entering a magnetic field
with velocity v, making an angle 𝜃 with the direction of the field.

The velocity v can be resolved into two perpendicular components v cos 𝜃 and
v sin 𝜃
The force acting on the charged particle due to the component of velocity v cos
𝜃 acting along the direction of the field is zero. So, the charged particle
continues to move along the direction of the field with a speed of v cos 𝜃.
Due to the perpendicular component of velocity v sin 𝜃, the charged particle
experiences a force F = qv B. This force makes the particle move along the
circular path in the YZ plane.
𝑚(𝑣sin 𝜃)2
q v B sin 𝜃 = 𝑟
𝑚𝑣𝑠𝑖𝑛𝜃
r= 𝑞𝐵
2𝜋𝑟 2𝜋𝑚
T =𝑣𝑠𝑖𝑛𝜃 = 𝑞𝐵
Therefore under the combined effect of two component of velocities the
charged particle in magnetic field will cover linear path as well as circular path.
Hence the resultant path will be a helix with its axis along B.
Pitch
The linear distance travelled by the charged particle in the direction of the
magnetic field during its period of revolution is called pitch of the helical path.
2𝜋𝑚
Pitch = v cos𝜃 × T = v cos𝜃 × 𝑞𝐵
Motion in combined electric and magnetic fields (Velocity selector)

A charge q is moving with velocity v in presence of electric and magnetic field

then Lorentz force
F = q 𝐸⃗ + q ⃗⃗⃗⃗ ⃗)
(𝑣 × 𝐵
5
Considser a situation in which v, E and B are mutually perpendicular to each
other.
𝑣 = v 𝑖̂
𝐸⃗ = 𝐸𝑗̂
𝐵⃗ = 𝐵𝑘̂
Therefore,
Force F = q 𝐸⃗ + q ⃗⃗⃗⃗
(𝑣 × 𝐵 ⃗)
= q 𝐸𝑗̂ + 𝑞 (v 𝑖̂ × 𝐵𝑘̂ )
= q 𝐸𝑗̂ - 𝑗̂ q v B
Thus, electric and magnetic forces are in opposite directions. If we adjust the
value E and B such that magnitudes of two forces are equal, then the total
force on the charge is zero and the charge will move in the fields undeflected.
i.e., q 𝐸𝑗̂ - 𝑗̂ q v B = 0
𝑬
v=𝑩
this condition can be used to select charged particles of a particular velocity
out of a beam containing charges moving with different speeds. The crossed E
and B fields therefore serve as a velocity selector. Only the particles with
velocities
𝑬
passes undeflected through the region of crossed fields.
𝑩
Force on a current carrying conductor placed in a magnetic field.
Consider a wire of uniform cross sectional area A and length L.

Let there are n electrons per unit volume and the drift velocity of the electrons
be vd.
The total number of electrons in it = nAL
Force acting on these electrons F = -(nALe) Vd B Sin𝜃
Since I = -nAeVd (- sign because current and drift velocity are in opposite
direction)
F =I L B Sin𝜃
𝐹 = I 𝐿⃗ × 𝐵
⃗
Here 𝐿⃗ is a vector of mangitude L and with directions towards the current I.
The directions of the force is perpendicular to the plane containing B and L.

6
If 𝜃 = 0 𝑜𝑟 180 𝑡ℎ𝑒𝑛 𝐹 = 0
If 𝜃 = 90 𝑡ℎ𝑒𝑛 𝐹 =ILB
The direction of force is give be fleming left hand rule.
Biot Savart law
This is an experimental law predicted by Biot and Savart which gives the
relation between current and Magnetic field it produces.
This law deals with the magnetic field at a point due to a small current
element.
Current element is a part of any conductor carrying current and can be
represented as product of current and length of a very small segment
carrying current. Current element is a vector quantity and its direction is
tangent to the element and is acting in the direction of current flow
Let us consider a small element of length dl of a conductor XY carrying a
current I.

Let 𝑟 be the position vector of the point P from the current element I ⃗⃗⃗⃗
𝑑𝑙 and 𝜃
⃗⃗⃗ and 𝑟
be the angle 𝑑𝑙
According to Biot Savart,
The magnitude of magnetic field dB at a point P due to current element
i. dB ∝ 𝐼
ii. dB ∝ 𝑑𝑙
iii. dB ∝ 𝑠𝑖𝑛𝜃
1
iv. dB ∝ 𝑟 2
combining
𝐼𝑑𝑙𝑠𝑖𝑛𝜃
dB ∝
𝑟2
𝐼𝑑𝑙𝑠𝑖𝑛𝜃
dB= 𝑘 𝑟2
where, 𝜇0 is the absolute magnetic permeability in free space.
𝜇0 = 4𝜋 × 10−7 𝑇𝑚𝐴−1
𝜇
k = 0 = 10−7 in SI units
4𝜋
k = 1 in CGS units

7
 𝜇 𝐼𝑑𝑙𝑠𝑖𝑛𝜃
dB = 4𝜋0 𝑟2
In vector form
⃗⃗⃗ × 𝑟
𝜇0 𝐼𝑑𝑙
⃗⃗⃗⃗⃗
𝑑𝐵 =
4𝜋 𝑟 3
The direction of dB is given by Right hand rule.
Similarities between Biot Savart law for magnetic field and coulombs law for
electric field.
Similarities
• Both are long range since both depends inversely on the square of
distances from the sources and point of observation.
• Both fields obey super position principle.
• Magnetic field is linear in the source 𝐼𝑑𝑙
⃗⃗⃗ just as the electric field is linear
in its source the electric charge q.
Dissimilarities
• Electrostatic field is produced by a scalar source (electric charge) and
magnetic field is produced by vector source 𝐼𝑑𝑙⃗⃗⃗ .
• Electric field is along the direction of displacement vector r, but B is
acting perpendicular to the plane containing the current element and
displacement vector r.
• There is an angle dependence in Biot Savart law which is not present in
coulombs law.
Magnetic field on the axis of a circular current loop
Consider a circular coil of radius a with centre O carrying current I. Let P
be a point on the axis of the circular coil at a distance x from the centre
O.
Now consider two small elements of the coil each of length dl at A and B
which are situated diametrically opposite edges.

According to Biot Savart Law the magnetic field at P due to the current
element I dl at A is given by

8
 𝜇 𝐼𝑑𝑙𝑠𝑖𝑛90
dB = 4𝜋0 ( dl ⊥ r) 𝜃 = 90
𝑟2
𝜇 𝐼𝑑𝑙
dB = 4𝜋0 --------------- 1
𝑎2 +𝑥 2
the direction of dB is in the plane of paper and is perpendicular to dl and
r
similarly
the magnetic field at P due to the current element I dl at B is given by
𝜇 𝐼𝑑𝑙
dB = 4𝜋0 ---------------2
𝑎2 +𝑥 2
dB can be resolved into two components dB sin𝜙 along axis and dB cos𝜙
perpendicular to the axis.
For any two diametrically opposite elements of the loop, the components
perpendicular to the axis will be cancelled and the components along the
axis will get add up.
Total magnetic field at the point P is in the direction OP
𝜇 𝐼𝑑𝑙 𝜇 𝐼
B = ∫ dB sin𝜙 = ∫ 4𝜋0 sin𝜙 = 4𝜋0 sin𝜙 ∫ 𝑑𝑙
𝑎2 +𝑥 2 𝑎2 +𝑥 2
Substitute,
𝑎
sin𝜙 = 2 and ∫ 𝑑𝑙 = 2𝜋𝑎
√𝑎 +𝑥 2
then,
𝜇 𝐼 𝑎
B = 4𝜋0 × × 2𝜋𝑎
𝑎2 +𝑥 2 √𝑎2 +𝑥 2
𝜇0 2𝜋𝑎2 𝐼
B= acts along PX
4𝜋 (𝑎2 +𝑥 2)3⁄2
If there are n turns then,
𝜇 2𝜋𝑛𝑎2 𝐼
B = 4𝜋0 3⁄
(𝑎2 +𝑥 2) 2

Special cases
i. When the point P lies at the centre then x = 0
𝜇 2𝜋𝑛𝐼 𝜇0𝑛𝐼
B = 4𝜋0 𝑎
= 2𝑎
ii. When the point P lies far away from the centre x >>a
𝜇 2𝜋𝑛𝑎2 𝐼
B = 4𝜋0 𝑥3
Note:
The current loop can be regarded as a magnetic dipole which produces
its magnetic field and the magnetic dipole moment of the current loop is
equal to the product of current, number of turns and area of current
loop.
Magnetic dipole moment, m = n I A
SI unit Am2

9
 Ampere’s circuital law
It states that the line integral of magnetic field around a closed path in
vacuum is equal to 𝜇0 times the total current I passing through the
closed path.
⃗⃗⃗ = 𝜇0 𝐼
⃗ . 𝑑𝑙
∮𝐵
Note: While finding magnetic field due to current using Amperes
circuital law, we have to choose a closed loop called an amperian loop
such that at each point of the loop either
i. B is tangential to the loop or
ii. B is a non-zero constant or
iii.B is normal to the loop or
iv. B vanishes
Magnetic field due to infinite long straight wire carrying current.

Consider an infinite long straight wire lying in the plane of the paper. Let
I be the current flowing through it. Let P be a point at a distance r from
the straight wire and B be the magnetic field at P.
Consider an amperian loop as a circle of radius r perpendicular to the
plane of the paper with centre on wire such that the point lies on the
loop. The magnitude of B is same at all points on this loop and B at the
point P will be tangential to the circumference of the loop.
⃗ . ⃗⃗⃗
∮𝐵 𝑑𝑙 = ∮ 𝐵𝑑𝑙 = 𝐵 ∮ 𝑑𝑙 = 𝐵2𝜋𝑟
Using amperes circuital law
⃗ . ⃗⃗⃗
∮𝐵 𝑑𝑙 = 𝜇0 𝐼
𝐵2𝜋𝑟 = 𝜇0 𝐼
𝜇0 𝐼
𝐵=
2𝜋𝑟
This relation tells that

10
 • Magnitude of magnetic field at every point on a circle of radius r
with wire along the axis is same.
• The magnetic field due to a current carrying straight conductor
has a cylindrical symmetry.
• The direction of field at every point on the circle is tangential.
• The magnetic field lines form concentric circles.
• Theoretical proof to oersted experiment.
Magnetic field due to current through a very long circular cylinder
Consider an infinite long cylinder of radius R. Let I be the current passing
through the cylinder. The magnetic field are perpendicular to the length
of the cylinder.

Case 1 (Point P lying outside the cylinder)

Let r be the perpendicular distance of point P from the axis of the
cylinder.
The magnetic field is acting tangential to the field line at P.
Here B and dl are acting in the same direction
Applying Amperes circuital law
⃗⃗⃗ = 𝜇0 𝐼
⃗ . 𝑑𝑙
∮𝐵
𝐵2𝜋𝑟 = 𝜇0 𝐼
𝜇0 𝐼
𝐵=
2𝜋𝑟
Case 2 (Point P is inside the cylinder)
Here we may have two possibilities
i. If the current is only along the surface of cylinder, then current
through the closed loop L is zero
using Ampere circuital law B = 0
ii. If the current is uniformly distributed throughout the cross section of
the conductor, then the current through closed path L is given by
11
 𝐼 𝐼𝑟 2
I’ = 𝜋𝑅2 𝜋𝑟 2 = 𝑅2
Applying Ampere circuital law,
⃗⃗⃗ = 𝜇0 𝐼 ′
⃗ . 𝑑𝑙
∮𝐵
𝐼𝑟 2
B (2𝜋𝑟) = 𝜇0 𝑅2
𝜇0 𝐼𝑟
B= 2𝜋𝑅2
B∝ 𝑟

So, the magnetic field is maximum for a point on the surface of solid
cylinder carrying current and is zero for a point on the axis of cylinder.
Magnetic field due to a Solenoid
A solenoid consists of an insulating long wire closely wound in the form
of a helix. Its length is very large compared to its diameter.

The current enters the plane of the paper at the points marked as ꚛ and
leaves the plane of paper at points marked as ꙩ.
When current is passed through the solenoid, each turn can be regarded
as a circular loop carrying current producing magnetic field.
At a point outside the solenoid the magnetic field due to the neighboring
loop oppose each other and adds up in the interior of the solenoid.
As a result, the effective magnetic field outside the solenoid becomes
weak, whereas the magnetic field in the interior of the solenoid is strong
and uniform, acting along the axis of the solenoid.

12
When the solenoid is more and more tightly packed, the magnetic field
due to current will become zero for a point outside near the middle of
the solenoid and uniform at a point in the interior of the solenoid.
Let n be the number of turns per unit length of solenoid and I be the
current flowing through the solenoid and the turns of the solenoid be
closely packed.

Now consider a rectangular amperian loop ABCD where AB = L

Total number of turns in length L = nL
Using Ampere circuital law
𝐵 𝐶 𝐷 𝐴
⃗ . ⃗⃗⃗
∮𝐵 ⃗ . ⃗⃗⃗
𝑑𝑙 = ∫ 𝐵 ⃗ . ⃗⃗⃗
𝑑𝑙 + ∫ 𝐵 ⃗ . ⃗⃗⃗
𝑑𝑙 + ∫ 𝐵 ⃗ . ⃗⃗⃗
𝑑𝑙 + ∫ 𝐵 𝑑𝑙
𝐴 𝐵 𝐶 𝐷
Field along CD is zero and as the path BC, AD are perpendicular to axis of
solenoid, the magnetic field along these path is zero.
𝐵
⃗ . ⃗⃗⃗
∮𝐵 ⃗ . ⃗⃗⃗
𝑑𝑙 = ∫ 𝐵 𝑑𝑙 = 𝐵𝐿 = 𝜇0 × 𝑡𝑜𝑡𝑎𝑙 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝐴𝐵𝐶𝐷
𝐴
= 𝜇0 𝐼nL
B = 𝜇0 𝑛𝐼
This relation gives the magnetic field at a point well inside the solenoid.
At a point near the end of the solenoid the magnetic field is found to be
𝜇0 𝑛𝐼
2
Force between two parallel current carrying conductor
Consider two infinite long straight conductors carrying currents 𝐼1 and
𝐼2 in the same direction. They are held parallel to each other at a
distance r.

13
Since magnetic field is produced due to current through each conductor,
therefore each conductor experience force.
The magnetic field produced by 𝐼1 at any point in the conductor C
𝜇0 𝐼1
𝐵1 =
2𝜋𝑟
The direction of this field is perpendicular to conductor C and point into
the plane of the paper.
This field exerts a force on the conductor C
𝜇0 𝐼1𝐼2 𝑙 𝜇0
𝐹2 = 𝐵1 𝐼2 𝑙 sin 90 = = 𝐼𝐼 𝑙
2𝜋𝑟 2𝜋𝑟 1 2
Force per unit length
𝐹2 𝜇0
F= 𝑙
= 2𝜋𝑟 𝐼1 𝐼2
According to Flemings left hand rule the direction of force is towards left
in the plane of the paper.
Similarly, an equal force is exerted on the wire A by the field produced
by the conductor C and the direction is opposite to 𝐹2
𝐹1 = −𝐹2
Which is in consistent with Newtons third law.
Thus,
Two parallel conductors carrying current in the same direction attract
each other.
Two parallel conductors carrying current in the opposite direction repel
each other
Standard definition for One Ampere
We have
𝜇0
F = 2𝜋𝑟 𝐼1 𝐼2 𝑙
When 𝐼1 = 𝐼2 = 𝑟 = 1
𝜇
F = 2𝜋0 = 2 × 10−7 𝑁𝑚−1
One Ampere is that current which when flows through two parallel
wires separated by one metre in vacuum produces a force of
2 × 𝟏𝟎−𝟕 𝑵 on unit length of the wire.
Torque on a rectangular current loop in a uniform magnetic field
Consider a rectangular coil PQRS suspended in uniform magnetic field B
with its axis perpendicular to the field.
Let I be the current PQ = SR = b and PS = RS = a
And 𝜃 be the angle between the direction of the field and normal drawn
to the plane of the coil.

14
 Force on arm PS = I a B Sin(90 − 𝜃) = I a B Cos 𝜃
Force on arm QR = I a B Sin(90 − 𝜃) = I a B Cos 𝜃
These two collinear forces are equal and opposite so they cancel
each other.
Force on arm RS = I b B Sin 90 = Ib B
The direction of the force is points into the plane of paper.
Force on arm PQ = I b B Sin 90 = Ib B
The direction of the force is points towards the reader.
These two forces are equal and opposite and forms a couple

Torque 𝜏 = 𝑓𝑜𝑟𝑐𝑒 × 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒

= F× a sin 𝜃 = I b B × a sin 𝜃 =I AB sin 𝜃
= 𝑚𝐵 sin 𝜃
=𝑚×𝐵
𝜏=𝑚 ⃗
⃗⃗ × 𝐵
The direction of torque is such that it rotates the loop in clockwise
direction.
Stable equilibrium
When 𝜃 = 0
𝜏=0
Unstable equilibrium
When 𝜃 = 180
𝜏=0

15
 Circular current loop as a magnetic dipole

Magnetic field on the axis of a circular current loop

𝜇 2𝜋𝑛𝑎2 𝐼
B = 4𝜋0 3⁄
(𝑎2 +𝑥 2) 2

And its direction is along the axis and give by right hand thumb rule.
𝜇 2𝜋𝑛𝑎2 𝐼
When x>>a, we can write B = 4𝜋0 𝑥3
2
Area of the loop A = 𝜋𝑎
𝜇0 𝑛𝐼𝐴 𝜇0 𝑚
B = =
2𝜋𝑥 3 2𝜋𝑥 3

𝜇0 2𝑚
𝐵=
4𝜋 𝑥 3
This expression is similar to an expression obtained for electric dipole
1 2𝑃
E = 4𝜋𝜖
0 𝑥3
Thus, a current loop is equivalent to a magnetic dipole of dipole moment
(m)
Thus
A current loop
i. Produce magnetic field
ii. Behaves like magnetic dipole
iii. Subject to torque like a magnetic dipole.
This led Ampere to suggest that magnetism is due to
circulatory currents.
Magnetic dipole moment of a revolving electrons
According to Bohr model of hydrogen atom, the negative electron is revolving
around a central positive charged nucleus. The uniform circular motion of
electron constitutes a current.

16
Since the electron is revolving in anticlockwise direction the current is in
clock wise direction. The magnetic field developed will be given by right
hand thumb rule and is pointing into the plane of paper.
𝑒
Current I = 𝑇 , where T is the time period of revolution.
2𝜋𝑟
Time period, T = 𝑣
𝑒𝑣
Current I = 2𝜋𝑟
Magnetic dipole moment associated with the circulating current, 𝜇𝑙 = IA
= I 𝜋𝑟 2
𝑒𝑣 𝑒𝑣𝑟
𝜇𝑙 = 2𝜋𝑟 × 𝜋𝑟 2 = 2

𝑒𝑣𝑟 × 𝑚𝑒 𝑒𝑚𝑒 𝑣𝑟 𝑒𝑙
𝜇𝑙 = = =
2𝑚𝑒 2𝑚𝑒 2𝑚𝑒
Where l is orbital angular momentum
𝑒𝑙
𝜇𝑙 = −
2𝑚𝑒
Negative sign indicates that angular momentum of electron is
opposite to magnetic moment
If we consider a particle with charge +e then
𝑒𝑙
𝜇𝑙 =
2𝑚𝑒
Here angular moment and magnetic moment is in same direction.
𝜇𝑙 𝑒
=
𝑙 2𝑚𝑒
𝝁
This ratio 𝒍𝒍 is known as gyromagnetic ratio and its is a constant. Its
value is 8.8 × 𝟏𝟎𝟏𝟎 𝑪/𝒌𝒈 for an electron.
Again
Magnetic moment,
𝑒𝑙
𝜇𝑙 =
2𝑚𝑒
From Bohr’s quantization condition
𝑛ℎ
𝑙= 𝑤ℎ𝑒𝑟𝑒 𝑛 = 1,2,3, …
2𝜋
For n = 1
Magnetic moment
𝑒 ℎ 𝑒ℎ
𝜇𝑙 = =
2𝑚𝑒 2𝜋 4𝜋𝑚𝑒

17
 𝒆𝒉
𝝁𝒍 =
𝟒𝝅𝒎𝒆
Value of planks constant h =6.65 x 10-34 J Hz-1
Mass of an electron = 9.1 × 10-31 kilograms
This is called Bohr magneton and its value is
9.27 × 𝟏𝟎−𝟐𝟕 Am2
𝜇𝑙 is the magnetic moment associated with a charge in uniform circular
motion and hence it is known as orbital magnetic moment.
Moving coil galvanometer
It is an instrument used to detect current.

Principle: It is based on the fact that a current carrying coil placed in a

magnetic field experiences a torque.
Construction: It consist of a coil with many turns free to rotate about a
fixed axis in uniform radial magnetic field. There is a cylindrical soft iron
core which not only makes the field radial, but also increases the
strength of the magnetic field.
When current flows through the coil a torque act on it
𝜏 = 𝑁𝐼𝐴𝐵𝑠𝑖𝑛𝜃 = 𝑁𝐼𝐴𝐵
𝑠𝑖𝑛𝜃 = sin 90 = 1
since normal to the plane of the coil remains perpendicular to the field
at all positions.
This torque deflects the coil through an angle 𝛼. A restoring torque will
be set up in the coil due to the elasticity.
Restoring torque is proportional to 𝛼
Restoring torque = k 𝛼
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Where k is a constant called torsion constant.
In equilibrium
Restoring torque = deflecting torque
k 𝛼 = 𝑁𝐼𝐴𝐵
𝑁𝐴𝐵
𝛼= I
𝐾
𝛼 ∝𝐼
Thus, deflection produced in the galvanometer is proportional to the
current flowing through it.
𝑘
I = 𝑁𝐴𝐵 𝛼
𝑘
The constant 𝑁𝐴𝐵 is called the galvanometer constant or current
reduction constant.
Figure of merit of a galvanometer
Figure of merit of a galvanometer is defined as the current which
produces a deflection of one scale division in the galvanometer.
𝐼 𝑘
=
𝛼 𝑁𝐴𝐵
Sensitivity of a galvanometer
A galvanometer is said to be sensitive if it shows large scale deflection
even when a small current is passed through it or a small voltage is
applied across it.
Current sensitivity
Current sensitivity is defined as the deflection produced in the
galvanometer when a unit current flows through it.

𝛼 𝑁𝐴𝐵
𝐼𝑠 = =
𝐼 𝐾
Voltage sensitivity
Voltage sensitivity is defined as the deflection produced in the
galvanometer when a unit potential difference is applied across it ends.
𝛼 𝛼 𝑁𝐴𝐵
𝑉𝑠 = = =
𝑉 𝐼𝑅 𝐾𝑅
Note:
If the radial field were not present in a moving coil galvanometer, then
torque = 𝑁𝐼𝐴𝐵𝑠𝑖𝑛𝜃
1
and therefore, current would be proportional to
𝑠𝑖𝑛𝜃
the scale would be then be nonlinear and difficult to calibrate or to read
accurately.
19
 Conversion of galvanometer to ammeter
An ammeter is a low resistance galvanometer. It is used to measure the
current in a circuit in Amperes.
Galvanometer is a very sensitive device; it gives a full-scale deflection for
a current of the order of micro ampere. Also, for measuring current the
galvanometer has to be connected in series and as it has a large
resistance, this will change the value of the current in the circuit.
To overcome these difficulties a low resistance wire is connected in
parallel to the galvanometer.
The resistance connected in this way is known as shunt. Only a small
part of the total current passes through the galvanometer and remaining
current passes through the shunt.
The value of shunt resistance depends on the range of the current
required to be measured.

Let G = resistance of galvanometer

Ig= current with which galvanometer gives full scale deflection
S = shunt resistance
I - Ig = current through the shunt
As the galvanometer and shunt are connected in parallel
Pot difference across G = pot difference across shunt
𝐼𝑔 𝐺 = (𝐼 − 𝐼𝑔 )𝑆
𝐼𝑔 𝐺
𝑆=
𝐼 − 𝐼𝑔
The effective resistance of the converted galvanometer
𝐺𝑆
R = 𝐺+𝑆
An ideal ammeter has zero resistance
Conversion of galvanometer into voltmeter
A voltmeter is a high resistance galvanometer. It is used to measure the
potential difference between two points of a circuit in volts.
A galvanometer can be converted into a voltmeter by connecting a high
resistance is series with the galvanometer, so that, it draws a small current.
20
Otherwise, the voltage measurement will disturb the original setup by an large
amount which is very large.

Total resistance of the circuit = R + G

By ohms law,
𝑉
𝐼𝑔 =
𝑅+𝐺
𝑉
R= −𝐺
𝐼𝑔
Effective resistance of converted voltmeter
Rv = R + G
---------------------------------------------------------

21