1. 4.Moving Charges and Magnetism
Hans – Oersted Experiment:
1. Christian Oersted discovered that the conductor carrying current produces magnetic
field around it. This effect is called magnetic effect of electric current.
2. Danish physicist Hans Christian Oersted noticed that a current in a straight wire
caused a deflection in a nearby magnetic compass needle.
Reversing the direction of the current reverses the orientation of the needle . The
deflection increases on increasing the current or bringing the needle closer to the wire.
Iron filings sprinkled around the wire arrange themselves in concentric circles
Oersted concluded that “moving charges or currents produced a magnetic field in the
surrounding space. Unlike a stationary charge which only produces electric field.”
3. The direction of magnetic field lines is given by amperes right hand classed rule.
ELECTRICITY AND MAGNETISM WERE UNIFIED BY JAMES MAXWELL WHO THEN REALISED
THAT LIGHT WAS ELECTROMAGNETIC WAVES.
(a) The current emerges out of the plane of the paper, The orientation of the magnetic needles
(b)The current moves into the plane of the paper. around current carrying conductor
Sign convention:
 A current or a field (electric or magnetic) emerging out of the plane of the paper is
depicted by a dot ( )
 A current or a field going into the plane of the paper is depicted by a cross (⊗)
Right Hand Thumb Rule: If the conductor carrying current held in right hand such that
thumb indicates the direction of current in the conductor, then direction of remaining
fingers gives direction of magnetic field.
a b
with the wire as the .
Page 69

2. 4.Moving Charges and Magnetism
OBTAIN AN EXPRESSION FOR FORCE ON CHARGE MOVING IN UNIFORM
MAGNETIC FIELD.
The force acting on charge ‘q’ moving with velocity ‘v’ making an angle θ with
uniform magnetic field ‘B’ is
1. Directly proportional to the magnitude of charge(q).
2. Directly proportional to the velocity of charge (v).
3. Directly proportional to magnetic field (B).
4. Directly proportional to sign of angle between ‘v’ & ‘B’
F ∝ qvBsinθ
F = qvBsinθ
In vector form 𝐅
⃗ = 𝐪(𝐯
⃗⃗ × 𝐁
⃗⃗⃗).
“The direction of force is given by right hand palm rule for positive charge and
left hand palm rule for negative charge. Stretch all fingers of right/left hand in the
direction of charge particle and curl all fingers in the direction of magnetic field except
thumb, then thumb represents the direction of force.”
Note:
 The force acting on charge is maximum when it is moving ⊥er to field direction (θ
= 900).
 The force on charge is zero(minimum) when charge is moving parallel or
antiparallel to magnetic field.( θ = 00 or 1800).
 Charge is at rest (v=0) then F = 0.
 Uncharged particle (q=0) like neutron then F = 0.
What is S.I unit of magnetic field?
Ans:Tesla (T), vector quantity, [MT2A-1]
Define Tesla: If a charge is 1 coulomb is moving with velocity 1ms-1 perpendicular to
magnetic field experiences a force of 1 Newton then the magnetic field is 1 Tesla.
CGS unit of magnetic field is Gauss (1T = 10-4G).
LORENTZ FORCE: Lorentz force is defined as the combination of the magnetic and
electric force on a point charge due to electromagnetic fields.
Force due to magnetic field, FB = qvB
Force due to electric field, FE = qE
𝑭
⃗
⃗⃗ = 𝑭𝑩 + 𝑭𝑬
𝐹
⃗ = 𝑞𝑣𝐵 + 𝑞𝐸
𝑭
⃗
⃗⃗ = 𝒒(𝒗
⃗
⃗⃗ × 𝑩
⃗⃗⃗ + 𝑬
⃗⃗⃗) vector form of Lorentz force.
Page 70

3. 4.Moving Charges and Magnetism
𝑭 = 𝑰𝑩𝒍𝒔𝒊𝒏𝜽
Where,
 F is the force acting on the particle
 q is the electric charge of the particle
 v is the velocity
 E is the external electric field
 B is the magnetic field
Applications of Lorentz force:
Cyclotrons and other particle accelerators use Lorentz force.
OBTAIN AN EXPRESSION FOR FORCE ON CURRENT CARRYING CONDUCTOR
PLACED IN MAGNETIC FIELD
Consider a conductor of length l, area of cross section ‘A’ carrying a current
‘I’ is placed ⊥er to uniform magnetic field ‘B’. Let ‘n’ be the free electrons in the
conductor. Then total no. of free electrons in the conductor is ‘nAl’. The total charge of all
electrons is nAlq. The force acting on all the charges is
F = nAlq (Vd x B) Vd =drift velocity
F = (AnqVd)l B
F = IBl Where I = AnqVd (From chapter 3)
For any arbitrary angle ‘θ’
Note: When the force on conductor carrying current placed in the uniform magnetic
field I) maximum II) minimum
I) Fmax: When conductor is perpendicular to magnetic field, θ = 900
II) Fmin: When conductor is parallel or antiparallel to magnetic field. ( θ = 00 or 1800).
Solve Numerical number 4.1 from NCERT text book
OBTAIN AN EXPRESSION FOR RADIUS OF CIRCULAR PATH OF CHARGE MOVING IN
UNIFORM MAGNETIC FIELD. Consider a charge q moving with velocity
‘v’ ⊥er to uniform magnetic field ‘B’
Charge experience force
FB= Bqvsinθ
FB = Bqv (θ = 900)
As a result charge perform circular motion.
The centripetal force on charge is
FC =
𝒎𝒗𝟐
𝒓
page 71

4. 4.Moving Charges and Magnetism
For equilibrium ,force on charge due o magnetic field is balanced by centripetal force.
Hence, FB= FC
Bqv =
𝑚𝑣2
𝑟
Bq =
𝑚𝑣
𝑟
r =
𝒎𝒗
𝑩𝒒
This is expression for radius of circular path.
Note: (NEET/CET) Other forms of radius , r =
𝒎𝑽
𝑩𝒒
=
𝑷
𝑩𝒒
=
√𝟐𝒎𝑲
𝑩𝒒
=
𝟏
𝑩
√
𝟐𝒎𝑽
𝒒
Time period(T): The time required to complete one rotation for a charge particle.
Time period T =
𝟐𝝅𝒎
𝑩𝒒
Frequency(f): Number of rotation in one second.
Frequency f =
𝑩𝒒
𝟐𝝅𝒎
Time period (or frequency) is independent of speed of particle and radius of the orbit
and depends only on the field B and the nature
HELICAL MOTION/SPRING LIKE MOTION
Motion of charged particle in uniform magnetic field at arbitrary angle ‘θ’
Consider a charged particle makes an arbitrary angle ‘θ’ with the magnetic field
(i.e. θ ≠ 900 or 00) as shown in figure.
The velocity ‘v’ makes angle ‘θ’ with B which is along the x – axis. The velocity
vector splits into two components.
Page 72

5. 4.Moving Charges and Magnetism
1) vcosθ along horizontal ,responsible for forward motion
2) vsinθ along vertical, responsible for vertical circular motion
Due to these two component, charge particle move vertical as well as horizontal at
the same time. Hence the nature of this motion is spiral like known as Helix.
The radius(r) and pitch(P) of the helical motion is given by
r =
𝒎𝒗𝒔𝒊𝒏𝜽
𝒒𝑩
P =
𝟐𝝅𝒎𝑽𝒄𝒐𝒔𝜽
𝑩𝒒
Solve Numerical number 4.3 from NCERT text book
CYCLOTRON
The cyclotron is a machine to accelerate charged particles or ions (like, α-
particles, deutrons etc.) to high energies. It was invented by E.O. Lawrence and M.S.
Livingston.
Principle: Cyclotron uses the fact that the frequency of revolution of the charged
particle in a magnetic field is independent of its energy i.e kinetic energy.
Construction & Working:
WORKING: During +ve (positive) half cycle of AC. D1 becomes +ve and D2 negative (-
ve) . Charged particles experiences an electrical force and accelerate towards D2 . Once
the charge particle reach into hallow dees ,the magnetic field, however, acts on the
particle and makes it go round in a circular path inside a dee. Every time the particle
moves from one dee to another it is acted upon by the electric field. This ensures that
It consist of two D shaped
hollow metal disc placed opposite with
gap, they are called dees. Dees D1 & D2
are connected to AC oscillator. A
uniform magnetic field is applied ⊥er to
the plane of dees. A charged particle to
be accelerated is produced at the
of dees in the gap.
Page 73

6. 4.Moving Charges and Magnetism
Solve Numerical number 4.4 from NCERT text
book
the particle is always accelerated by the electric field. Each time the acceleration
increases the energy of the particle.
Frequency of cyclotron is 𝒇𝒄 =
𝟐𝝅𝒎
𝑩𝒒
The kinetic energy of the ions is, 𝑲𝒎𝒂𝒙 =
𝒒𝟐𝑩𝟐𝑹𝟐
𝟐𝒎
What are the applications of cyclotron?
1) The accelerated charged particles are used to study structure of elements.
2) To implant ions into solids to change their properties.
3) To produce radioactive substances in hospitals for treatment.
4) The cyclotron is used to bombard nuclei with energetic particles, so accelerated by it,
and study the resulting nuclear reactions.
Assignment: "Write note on accelerators in India."
Describe the motion of charged particle in combined electric & magnetic field.
Consider a charge q moving with velocity 'v' along x – axis and an uniform electric
field ‘E’ is applied along y – axis and uniform magnetic field ‘B’ is along z – axis. There
for charge is acted by both forces. Then, total force on the charge is zero and the charge
will move in the fields undeflected. This happens when,
qE = qvB
E = vB
v =
𝑬
𝑩
[Velocity selector Formula]
The particle moving with velocity E/B do not experiences any force and travel
straight. Thus the particles of particular velocity can be separated from stream of particles
using crossed fields and is called velocity selector.
This method was used by J.J. Thomson to calculate e/m (charge ratio to the mass
of e - ). It is also used in mass spectrometer.
Page 74

7. 4.Moving Charges and Magnetism
MAGNETIC FIELD DUE TO A CURRENT ELEMENT,BIOT-SAVART LAW
I. Directly proportional to current(I) through element.
II. Directly proportional to length of element(dl).
III. Directly proportional to sinθ.
IV. Inversely proportional to square of distance( 𝑟2
)
dB ∝
𝐼𝑑𝑙𝑠𝑖𝑛𝜃
𝑟2
dB = (
𝜇0
4𝜋
)
𝐼𝑑𝑙𝑠𝑖𝑛𝜃
𝑟2
where (
𝜇0
4𝜋
) is constant of proportionality
µo ---is called permeability of free space(vacuum) and
𝜇0
4𝜋
= 10-7 TmA-1
Biot-Savart law in vector: 𝒅𝑩
⃗⃗⃗⃗⃗⃗⃗ = (
𝝁𝟎
𝟒𝝅
)
𝑰(𝒅𝒍
⃗⃗⃗⃗⃗×𝒓
⃗⃗)
𝒓𝟑
⃗⃗⃗⃗⃗
Similarities and differences between Biot-Savart law and Coulomb’s Law
 The current element produces a magnetic field, whereas a point charge produces
an electric field.
 The magnitude of magnetic field varies as the inverse square of the distance from
the current element, as does the electric field due to a point charge
 The electric field created by a point charge is radial, but the magnetic field created
by a current element is perpendicular to both the length element
Consider an element of length
dl carrying current ‘I’. Let ‘P’
be a point at distance ‘r’ from
According to Biot-Savart law
magnetic field 'dB' at ‘P’ due
to current element is
of current element.
Page 75

8. 4.Moving Charges and Magnetism
The direction of magnetic field is determined with the help of the following simple
laws :
(1) MAXWELL’S CORK SCREW RULE
According to this rule, if we imagine a right handed screw placed along the current
carrying linear conductor, be rotated such that the screw moves in the direction of flow
of current, then the direction of rotation of the thumb gives the direction of magnetic
lines of force.
(2) RIGHT HAND THUMB RULE
According to this rule if a current carrying conductor is held in the right hand such
that the thumb of the hand represents the direction of current flow, then the direction of
folding fingers will represent the direction of magnetic lines of force.
(3) RIGHT HAND THUMB RULE OF CIRCULAR CURRENTS
According to this rule if the direction of current in circular conducting coil is in the
direction of folding fingers of right hand, then the direction of magnetic field will be in
the direction of stretched thumb.
Page 76

9. 4.Moving Charges and Magnetism
OBTAIN AN EXPRESSION FOR MAGNETIC FIELD A POINT ON AXIS OF CURRENT
CARRYING CIRCULAR LOOP USING BIOT-SAVART LAW.(5marks)
Consider a circular loop of radius ‘R’ carrying current ‘I’. Let 'P' be a point on axis
at of length dl.
According to Biot-Savart law Magnetic field at P due to current element is
dB = (
𝜇0
4𝜋
)
𝐼𝑑𝑙𝑠𝑖𝑛𝜃
𝑟2
from figure, 𝑥2
+ 𝑅2
= 𝑟2
& 𝑠𝑖𝑛𝜃 = 1
𝑑𝐵 = (
𝜇0
4𝜋
)
𝐼𝑑𝑙
(𝑥2+𝑅2)
Resolve dB into component. Vertical component due to diametrically opposite
elements get cancelled where as the component along x-axis 'dB cosθ' get added hence
total magnetic field due to loop is
𝑑𝐵𝑥=𝛴𝑑𝐵𝑐𝑜𝑠𝜃
But 𝑐𝑜𝑠𝜃 =
𝑅
(𝑥
2
+𝑅2
)
1
2
⁄
B = (
𝜇0
4𝜋
)
𝐼
(𝑥2+𝑅2)
𝑅
(𝑥2+𝑅2)
1
2
⁄
𝛴𝑑𝑙
The summation of elements dl over the loop yields 2ПR, the circumference of the loop.
Thus, the magnetic field at P due to entire circular loop is
B = (
𝜇0
4𝜋
)
𝐼𝑅
(𝑥2+𝑅2)
3
2
⁄
× 2𝜋𝑅
B = (
𝜇0
4𝜋
)
2π𝐼𝑅2
(𝑥2+𝑅2)
3
2
⁄
distance ‘x’ from ‘O’. Consider diametrically opposite element
Page 77

10. 4.Moving Charges and Magnetism
Solve Numerical number 4.7 from NCERT text
book
𝐵 =
μ0𝐼𝑅2
2(𝑥2+𝑅2)
3
2
⁄
Note: B =
𝝁𝒐𝑰
𝟐𝑹
2) If There are ‘N’ number of turns in coil then
B =
𝝁𝒐𝑰𝑵
𝟐𝑹
Note: Ratio of Bcentre and Baxis ,
𝑩𝒄𝒆𝒏𝒕𝒓𝒆
𝑩𝒂𝒙𝒊𝒔
= (𝟏 +
𝒙𝟐
𝑹𝟐
)
𝟑
𝟐
⁄
AMPERES CIRCUTAL LAW
State and explains Amperes circutal law.
Statement: Line integral of 'Bdl' for a closed surface having current through it is ′μo′
times the total current(I) through the surface.
∮ 𝐵𝑑𝑙 = 𝜇𝑜𝐼
1) Using Amperes circutal law obtain an expression for magnetic field due to
straight infinite length wire carrying current.(3marks)
Consider an infinitely long straight wire
carrying current ‘I’. Let ‘P’ be a point at
distance r from wire. Imagine a circle of
amperian loop. Amperian loop has current
‘I’ through it by amperes circutal law the
magnetic field at distance ‘r’ from conductor
is
∮ 𝐵𝑑𝑙 = 𝜇𝑜𝐼
B. 2πr = 𝜇𝑜𝐼
B =
𝜇𝑜𝐼
2𝜋𝑟
radius ‘r’ having on wire as an
1) Magnetic field at of circular current loop i.e x=0,
Page 78

11. 4.Moving Charges and Magnetism
THE SOLENOID AND THE TOROID
The solenoid and the toroid are two pieces of equipment which generate
magnetic fields.
The television uses the solenoid to generate magnetic fields needed.
* What is solenoid: Solenoid is a helical cell of N turns of insulated wire wound closely.
* What is an ideal Solenoid: If length of solenoid is very much longer than its radius,
then it is called ideal solenoid (l >>>5r).
* How is magnetic field inside given solenoid made strong: It is done by inserting
laminated Core inside the solenoid.
Obtain expression magnetic field due to solenoid carrying current.(3marks)
Let n be number of turns per unit length of solenoid carrying ‘I’ for ideal
solenoid (large length). The field outside the solenoid approaches zero. We shall
assume that the field outside is zero. The field inside becomes everywhere parallel to
the axis.
To find magnetic field inside solenoid imagine a rectangular amperian loop ‘abcd’
of length ‘h’. The number of turns inside amperian loop is ‘nh’. Total current through
amperian loop is ‘nhI’ by ampere circuital loop
∮ 𝐵𝑑𝑙 = 𝜇𝑜𝐼
Bh = µonhI
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12. 4.Moving Charges and Magnetism
B = µonI
The direction of the field is given by the right-hand rule. The solenoid is commonly used
to obtain a uniform magnetic field
How the magnetic field along axial distance vary in case of a long solenoid.
Why solenoid tend to contract when a current passes through it: The parallel
current in adjacent turns create opposite magnetic poles which start attracting another
and solenoid tend to contract.
𝐵 =
𝜇0𝑛𝐼
2
Page 80

13. 4.Moving Charges and Magnetism
TOROID
What is toroid
Toroid is an endless solenoid bend inform of circle.
Using ampere circutal law expression for magnetic field inside toroid.
Toroid is an endless solenoid bend inform of circle. It is used to produce very
high magnetic field.
Let 'n' be the number of turns per unit length of toroid. ‘I’ is current through
toroid.
I) Magnetic field at a point in a free space inside toroid.
Let 'P' be a point at a distance 𝑟1
r1 element
through it. By ampere circutal law
∮ 𝐵𝑑𝑙 = µoI
B2π𝑟1 = µo(O)
B = O
Hence Magnetic field at a point in a free space inside toroid is ZERO.(1marks)
II) Magnetic field outside toroid: Let ‘Q’ be a point at distance r3
Imagine a circle of radius r3
inside amperian loop is equal to the current leaving the loop.
i.e. Net current I = 0
by ampere circuital law ∮ 𝐵𝑑𝑙 = µoI
B = O
Hence Magnetic field outside toroid is ZERO.
from ‘o’ of toroid. Imagine a circle of radius
and ‘o’ as an amperian loop. This amperian loop do not have current
from of toroid.
and ‘o’ has an amperian loop. The current entering
Page 81

14. 4.Moving Charges and Magnetism
Solve Numerical number 4.9 from NCERT text
book
III) Magnetic field inside toroid: Let ‘S’ be a point at distance r2 toroid. Imagine a circle
of radius r2
total no.of turns in toroid is N = nl
N = 2πrn
Total current entering the toroid is I(total)=2πrIn by ampere circuital law
∮ 𝐵𝑑𝑙 = µo(total) = 𝜇02𝜋𝑟𝐼𝑛
B. 2πr = µo2πrIn
B = µonI
FORCE BETWEEN TWO PARALLEL CURRENTS, THE AMPERE
We know that a current carrying conductor creates magnetic field around it
and magnetic field can exert a force. Hence if two conductor carrying current placed side
by side they experience a force on each other.
DERIVATION: Let two long parallel conductors 'a' and 'b' separated by a distance 'd'
and carrying (parallel) currents Ia and Ib, respectively.
The conductor ‘a’ produces, the same magnetic field Ba at all points along the
conductor ‘b’. The right-hand rule tells us that the direction of this field is downwards.
Its magnitude is given by Ampere’s circuital law,
𝐵𝑎 =
𝜇0𝐼𝑎
2𝜋𝑑
The conductor ‘b’ carrying a current Ib will experience force due to the field Ba.
The direction of this force is towards the conductor ‘a’. We label this force as Fba, the
force on a segment L of ‘b’ due to ‘a’. The magnitude of this force is given by
𝐹𝑏𝑎 = 𝐼𝑏𝐵𝑎𝐿 (Sinθ=1)
𝐹𝑏𝑎 =
𝜇0𝐼𝑎𝐼𝑏
2𝜋𝑑
𝐿
Similarly the force on ‘a’ due to ‘b’ is equal but opposite in direction.
Fba
Ba
and ‘o’ as amperian loop. This loop has current entering into it. The
Page 82

15. 4.Moving Charges and Magnetism
Solve Numerical number 4.10 from NCERT text
book
𝐹𝑏𝑎 = 𝐹𝑎𝑏 = 𝐹 =
𝜇0𝐼𝑎𝐼𝑏
2𝜋𝑑
L
Hence force per unit length of conductor is
𝐹
𝐿
=
𝜇0𝐼𝑎𝐼𝑏
2𝜋𝑑
The ampere is the value of that steady current which, when maintained in each
of the two very long, straight, parallel conductors of negligible cross-section, and placed
one metre apart in vacuum, would produce on each of these conductors a force equal to
2 × 10–7 newtons per metre of length.
NOTE:If conductors carries current in same direction, then force between them will be
attractive.
If conductor carries current in opposite direction, then force between them will be
repulsive.
Page 83

16. 4.Moving Charges and Magnetism
B
m
ϴ
𝒂
𝟐
⁄
𝒂
𝟐
⁄
𝒂
𝟐
𝐬𝐢𝐧 𝜽
𝑭𝟐
𝑭𝟏
TORQUE ON CURRENT LOOP, MAGNETIC DIPOLE
Consider a rectangular loop ABCD such that it carries a current of magnitude I.
This loop is placed in a magnetic field, it experiences a torque but no net force.
It quite similar to what an electric dipole experiences in a uniform electric field.
When the rectangular loop is placed such that the uniform magnetic field B is in the
plane of the loop. The field exerts no force on the two arms AD and BC of the loop.
The arm AB of the loop experience a force F1 ,Its magnitude is, F1 = IbB
Similarly, a force F2 on the arm CD, Its magnitude is F2 = IbB
Hence from above two equation we say, F1= F2
Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of
forces F1 and F2 hence loop rotates anticlockwise direction.
Net torque on rectangular loop is,
𝝉 = 𝒇𝒐𝒓𝒄𝒆 × 𝒑𝒆𝒓𝒑𝒆𝒏𝒅𝒊𝒄𝒖𝒍𝒂𝒓 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝜏 = 𝐹1 ×
𝑎
2
+ 𝐹2 ×
𝑎
2
𝜏 = 𝐼𝑏𝐵 ×
𝑎
2
+ 𝐼𝑏𝐵 ×
𝑎
2
𝜏 = 𝐼(𝑎𝑏)𝐵 but Area of loop (A) = ab
𝝉 = 𝑰𝑨𝑩
When the plane of the loop, is not along the magnetic field, but makes an angle(θ)
with it, then torque is given by
𝝉 = 𝑰𝑨𝑩 𝐬𝐢𝐧 𝜽
Page 84

17. 4.Moving Charges and Magnetism
For N number of turns 𝝉 = 𝑵𝑰𝑨𝑩 𝐬𝐢𝐧 𝜽
(i) 𝝉 is zero when θ = 0, i.e., when the plane of the coil is perpendicular to the field.
(ii) 𝝉 is maximum when θ= 90 , i.e., the plane of the coil is parallel to the field.
𝜏𝑚𝑎𝑥 =NBIA
Where 'm=IA' in above equation is called as "magnetic moment defined as the product
of area of the loop and current through the loop".
Unit of magnetic moment is --Am2 and Dimention formula is [AL2]
Vectorially torque is, 𝝉
⃗⃗ = 𝒎
⃗⃗⃗⃗ × 𝑩
⃗⃗⃗
Page 85

18. 4.Moving Charges and Magnetism
**THE MAGNETIC DIPOLE MOMENT OF A REVOLVING ELECTRON**
Obtain the expression for magnetic dipole moment of electron revolving around
nucleus.
Consider a electron of charge 'e' mass ‘m’ is revolving around nucleus in an orbit of
radius ‘ r’ current in orbit is I =
𝑒
𝑇
where T = Time period of electron.
T =
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
T =
2𝜋𝑟
𝑣
I =
𝑒
(
2𝜋𝑟
𝑣
)
=
𝑒𝑣
2𝜋𝑟
magnetic dipole moment of electron is
µe = IA =
𝑒𝑣
2𝜋𝑟
𝜋𝑟2
µe =
𝑒𝑣𝑟
2
-----------------(1)
Multiplied & divide eqn (1) by 'm'
µe =
𝑒𝑚𝑣𝑟
2𝑚
µe =
𝑒.𝑙
2𝑚
where l = mvr (angular momentum)
"The ratio of magnetic dipole moment of electron to that of angular momentum is
called as Gyromagnetic ratio" i.e.
𝝁𝒆
𝒍
=
𝒆
𝟐𝒎
= 8.8x1010 C /kg
by Bohr’s angular momentum condition l =
𝑛ℎ
2𝜋
µe =
𝑒
2𝑚
×
𝑛ℎ
2𝜋
µe =
𝒏𝒆𝒉
𝟒𝝅𝒎
This is magnetic dipole moment of electron in first (Ist) orbit of hydrogen atom for
n = 1 and is called Bohr magneton.(value of Bohr magneton µe = 9.27 x 10-27Am2)
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19. 4.Moving Charges and Magnetism
MOVING COIL GALVANOMETER
* What is moving coil galvanometer (MCG) ?
Moving coil galvanometer (MCG) is a device to
detect current.
Principle :-
When the current carrying coil is suspended in a
uniform magnetic field, then torque acts on coil
which deflect it. The deflection is directly
proportional to the magnitude of the current.
Construction :-
The galvanometer consists of a coil,
with many turns, free to rotate about a fixed
axis in a uniform radial magnetic field. There is
a cylindrical soft iron core which increases the
strength of the magnetic field. It consists a
rectangular coil of number of turns of insulated
wire pivoted between poles of magnet. It
rotates over semi-circular scale with central
‘0’(zero).
Working :-
When current is passed through it is deflected and experiences torque is given by
τ = nIAB
Where n = number of turns, I = current in coil A = Area of coil, B = Magnetic field
Due to deflection the spring is twisted and tends to bring the coil to original
position. Rotational effect of twist is called restoring couple whose moment is directly
proportional to deflection of coil.
𝜏 ∝ 𝜃
𝜏 = K 𝜃
Where K is couple per unit twist in equilibrium
nBIA = K 𝜃
I = (
𝐾
𝑛𝐵𝐴
) 𝜃
I ∝ 𝜃
The galvanometer cannot as such be used as an ammeter to measure
the value of the current in a given circuit. This is for two reasons:
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20. 4.Moving Charges and Magnetism
h
(i) Galvanometer is a very sensitive device, it gives a full-scale deflection for a current of
the order of μA.
(ii) For measuring currents, 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.
Explain the conversion of galvanometer to voltmeter.(3 marks)
Voltmeter is potential difference(PD) measuring device. It must be connected in
parallel in the circuit. It’s resistance must be high. Galvanometer cannot be used to
measure PD directly because its resistance is low.
Galvanometer can be converted into voltmeter by connecting high resistance in
series with it. Let ‘R’ be the high resistance connected in series with galvanometer and Ig
is current required to produce full scale deflection in galvanometer then, PD across
combination = Ig (R+G)
V = Ig(R+G)
𝑉
𝐼𝑔
= R+G
R=S =
𝑽
𝑰𝒈
– G
Voltage sensitivity: Deflection per unit volt is called voltage sensitivity of
galvanometer.
Unit: Division /volt
Page 88

21. 4.Moving Charges and Magnetism
Solve Numerical number 4.13 from NCERT text
book
Explain the conversion of Galvanometer into ammeter.( 3marks)
Ammeter is current measuring device. It must be connected in series. It’s
resistance must be low. Galvanometer cannot be used to measure current directly
because its resistance is not very low.
Galvanometer can be converted into ammeter by connecting low resistance in
‖el with it. Let ‘R’ be the resistance(shunt resistance) connected in ‖el with galvanometer
and Ig is current require to produce full scale deflection in galvanometer then,
PD across galvanometer = PD cross resistance
IgG = (I – Ig) S
S =
𝑰𝒈𝑮
𝑰−𝑰𝒈
Current sensitivity: Deflection per unit current is called current sensitivity.
Unit: division/A
Note: 1) Ideal voltmeter resistance is ′∞′ (infinite).
2) Ideal ammeter has resistance '0'(zero).
Distinguish b/w ammeter and voltmeter
Ammeter Voltmeter
1. It is current measuring device. 1. It is PD measuring device.
2.
Ammeter is a galvanometer with
low resistance in parallel
2.
Voltmeter is a galvanometer with high
resistance in series.
3. Ideal ammeter has zero resistance 3. Ideal voltmeter has resistance infinity
Page 89