Class 12 Physics Magnetic Field Notes

Class 12 Physics Magnetic Field Notes
UNIT 4
CLASS 12 PHYSICS • ELECTRICITY AND MAGNETISM

Magnetic Field

Chapter 16
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Oersted Discovery

Oersted experiment showing compass deflection reversing when current direction reverses N S Current in one direction Compass N S Current direction reversed Opposite deflection
Oersted experiment

Oersted discovered the magnetic effect of electric current. He found that when a wire carrying current is placed parallel to the compass needle, the needle gets deflected. On reversing the direction of current, the deflection of the needle was found to be in opposite direction.

Thus, an electric current produces magnetic effect in the space around the conductor.

Direction of Current and Magnetic Field

1. Right Hand Thumb Rule (Maxwell’s Right Hand Rule)

According to this rule, if a current carrying conductor is held in right hand, then thumb points the direction of current whereas curved fingers represent direction of magnetic field.

2. Right Hand Fist Rule (Right Hand Palm Fist Rule)

According to this rule, if the fingers of right hand are curved in the direction of flow of current in a circular coil, then the thumb points indicate the direction of magnetic field.

3. Fleming’s Left Hand Rule

According to this rule, if the thumb, fore finger and middle finger are stretched mutually perpendicular to each other such that the middle finger points the direction of motion of charge particle, the fore finger points the direction of magnetic field and thumb point gives the direction of force experienced by the conductor.
Clean schematic of right hand thumb rule and Fleming left hand rule Right hand thumb rule Current I Magnetic field B Fleming’s left hand rule Force F Field B Motion/current
Direction rules for magnetic field and force

Force on a Moving Charge in a Magnetic Field (Lorentz Force)

When a charged particle moves in a uniform magnetic field with certain velocity, it experiences a force in a direction perpendicular to both magnetic field and velocity. This force is called Lorentz force.
Velocity, magnetic field and Lorentz force directions for a moving charge +q Velocity v Magnetic field B Force F θ
Force on a moving charged particle in magnetic field

Let a charge q move with velocity v by making angle θ with the direction of magnetic field B. Experimentally the magnitude of force is found to be:

  1. directly proportional to charge: F ∝ q
  2. directly proportional to velocity of charge: F ∝ v
  3. directly proportional to strength of magnetic field: F ∝ B
  4. directly proportional to sine of angle between v and B: F ∝ sinθ
Combining the above relations:
F ∝ qvB sinθ
F = kqvB sinθ
For the expression used here, k = 1
F = qvB sinθ
In vector form:
F⃗ = q(v⃗ × B⃗)

Special Cases

When θ = 0° or 180°:
sinθ = 0
F = 0
When θ = 90°:
sinθ = 1
F = qvB
When charged particle is at rest, v = 0:
F = 0
For an electrically neutral particle, q = 0:
F = 0

This relation shows that the direction of magnetic force is perpendicular to the direction of magnetic field and velocity.

Force on a Current Carrying Conductor Placed in Magnetic Field

Current carrying conductor of length l making angle theta with uniform magnetic field Field B Current I θ l
Force on a current carrying conductor

Let a conductor of length l and cross-sectional area A be placed in magnetic field by making angle θ with magnetic field. Let I be the current through the conductor and vd be the drift velocity of free electron.

If n be the number of free electrons per unit volume, then the total number of free electrons in conductor is:

N = nAl

The force on each electron is:

F = e(vd × B)
Magnitude: F = evdB sinθ

Total force acting on the conductor:

F = NevdB sinθ
= nAle vdB sinθ

Since I = nAevd:

F = IlB sinθ
In vector form:
F⃗ = I(l⃗ × B⃗)

This shows that direction of force is perpendicular to magnetic field and length of conductor.

Special Cases

When θ = 0° or 180°:
F = 0
When θ = 90°:
F = IlB

Solved Numericals

Q.1 – Angle Made by a Current Carrying Conductor with Magnetic Field

A straight conductor of length 5 cm carries current of 1.5 A. The conductor experiences a magnetic force of 4.5×10−3 N when it is placed in a magnetic field of 0.9 tesla. What angle does the conductor make with B?

l = 5 cm = 5×10−2 m
I = 1.5 A
F = 4.5×10−3 N
B = 0.9 T

Using:

F = IlB sinθ
4.5×10−3 = (1.5)(5×10−2)(0.9) sinθ
sinθ = 6.66×10−2
θ = 3.8°

Q.2 – Current Required to Balance the Weight of a Rod

A straight horizontal rod X of mass 50 g and length 0.5 m is placed in a uniform horizontal magnetic field of 0.2 T perpendicular to X. Calculate the current in X if the force acting on it just balances its weight.

m = 50 g = 0.05 kg
l = 0.5 m
B = 0.2 T
θ = 90°

For balance:

mg = IlB sinθ
0.05(10) = I(0.5)(0.2) sin90°
I = (0.05×10)/(0.5×0.2)
I = 5 A

Biot-Savart’s Law

When current is passed through a conductor, magnetic field is produced around it. The magnitude of magnetic field produced is calculated by Biot-Savart’s law.
Biot-Savart geometry for a small current element dl and observation point P I dl P r θ
Biot-Savart law

Let XY be a conductor carrying current I. Suppose AB is a small element of conductor of length dl. There is a point P at distance r from centre of small element where magnetic field is to be found.

According to Biot-Savart’s law, magnetic field strength produced by the small element is:

  1. directly proportional to magnitude of current passed: dB ∝ I
  2. directly proportional to sine of angle between conductor and line joining P: dB ∝ sinθ
  3. inversely proportional to square of distance: dB ∝ 1/r²
  4. directly proportional to length of small element: dB ∝ dl
dB ∝ I dl sinθ / r²
dB = k I dl sinθ / r²
In SI unit:
k = μ₀/(4π)
μ₀ = 4π×10−7 H/m, permeability of free space
dB = (μ₀/4π)(I dl sinθ/r²)
The scan also notes the constants used in CGS electrostatic and CGS electromagnetic systems. The SI relation above is the main relation used in the following derivations.

Application of Biot-Savart’s Law: Magnetic Field at the Centre of a Current Carrying Circular Coil

Circular current carrying coil of radius r and magnetic field at its centre O r Current I Field at centre B
Magnetic field at the centre of a circular coil

Let us consider a circular coil of radius r carrying current I. Let dl be the length of a small element of coil.

dB = (μ₀/4π)(I dl sinθ/r²)
At the centre of circular coil, θ = 90°
dB = (μ₀/4π)(I dl/r²)

Total magnetic field due to whole coil:

B = ∮dB
= (μ₀I/4πr²)∮dl
∮dl = 2πr
B = μ₀I/(2r)

If the coil has n turns:

B = nμ₀I/(2r)

Magnetic Field due to an Infinitely Long Straight Current Carrying Conductor

Biot-Savart construction for magnetic field near an infinitely long straight conductor Conductor AB P a r dl
Magnetic field near an infinitely long straight conductor

Let AB be an infinitely long straight conductor carrying current I. Let dl be the small element and P be a point at distance a from O about which magnetic field is calculated.

dB = (μ₀/4π)(I dl sinθ/r²)

Using the geometry of the figure and integrating over the entire conductor:

B = (μ₀I/4πa) ∫−π/2π/2 cosα dα
= (μ₀I/4πa)[sinα]−π/2π/2
B = μ₀I/(2πa)

Magnetic Field on the Axis of Current Carrying Circular Coil

B = μ₀nIa² / [2(a² + x²)3/2]

Here a is radius of the circular coil, x is distance of point P from centre O and n is the number of turns.

Special Cases

1. If P lies at the centre of coil, x = 0:

B = μ₀nIa²/(2a³)
B = μ₀nI/(2a)

2. If P lies far away from O, x ≫ a:

(a² + x²) ≈ x²
B = μ₀nIa²/(2x³)

Magnetic Field on the Axis of a Long Solenoid

Long solenoid with observation point P on axis and end angles beta one and beta two P β₁ β₂ Long solenoid
Magnetic field due to a finite long solenoid

Let a long solenoid have n number of turns per unit length and carry current I. The magnetic field at point P due to the finite length of solenoid is:

B = (μ₀nI/2)(cosβ₁ − cosβ₂)

If the point P lies inside a solenoid of infinite length:

β₁ = 0°, β₂ = π
B = (μ₀nI/2)[1 − (−1)]
B = μ₀nI

Ampere’s Circuital Law

It states that the line integral of the magnetic field around any closed path is equal to μ₀ times the total current enclosed by the path.
∮ B⃗ · dl⃗ = μ₀I
Circular Amperian path around a long straight current carrying wire Current I r dl Closed Amperian path
Ampere’s circuital law

For a circular path of radius r around a long straight wire:

B = μ₀I/(2πr)
∮B⃗·dl⃗ = ∮B dl cos0
= B∮dl
= [μ₀I/(2πr)](2πr)
∮B⃗·dl⃗ = μ₀I

Applications of Ampere’s Circuital Law

1. Magnetic Field due to a Straight Current Carrying Conductor

∮B⃗·dl⃗ = μ₀I
B∮dl = μ₀I
B(2πr) = μ₀I
B = μ₀I/(2πr)

2. Magnetic Field due to Current Carrying Solenoid

Let a long solenoid have n number of turns per unit length and carry current I. The magnetic field outside the solenoid is very small, almost zero, and that inside is uniform.

For a rectangular Amperian path of length x inside solenoid:
Total turns enclosed = nx
Total current enclosed = nxI
∮B⃗·dl⃗ = μ₀(nxI)
Bx = μ₀nxI
B = μ₀nI

Magnetic Field due to Toroid

A toroid is a endless solenoid in the form of ring.
Toroid and circular Amperian path of radius r O r Toroid
Toroid

Let a toroid have n number of turns per unit length and current I be passed through each turn. The magnetic field outside toroid is zero.

For a circular path of radius r inside the toroid:

∮B⃗·dl⃗ = B(2πr)
Current enclosed = (n·2πr)I
According to Ampere’s circuital law:
B(2πr) = μ₀(n·2πr)I
B = μ₀nI

Hall Effect

When a magnetic field is applied to a current carrying conductor, a voltage is developed across a specimen in the direction perpendicular to both current and magnetic field. This effect is called Hall effect and the voltage thus produced is called Hall voltage.
Hall effect in a rectangular conductor showing current, magnetic field, charge separation and Hall voltage +++ +++ −−− −− Current I Field B Hall Vₕ d t
Hall effect

Suppose current I is flowing in x-axis and magnetic field is applied along z-axis. Then Hall voltage is produced across y-axis.

Let current I flow in a metal along positive x-direction. Free electrons drift with velocity vd in negative x-direction. When magnetic field is introduced, Lorentz force acts on electrons and bends them downward.

Due to downward deflection, electrons accumulate on lower surface of metal and produce net negative charge there. At the same time positive charge is produced on upper surface. This combination sets a downward electric field.

Magnetic force on electron: FB = evdB
Electric force: Fe = eE

When electric force and magnetic force are balanced:

eE = evdB
E = vdB   — (i)

Also, current density:

J = nvde
vd = J/(ne)   — (ii)

Hall voltage:

VH = Ed
E = VH/d   — (iii)

Using (i), (ii) and (iii):

VH/d = [J/(ne)]B
J = I/A and A = dt
n = IB/(VHte)
Therefore:
VH = IB/(net)

Torque on a Rectangular Coil in a Uniform Magnetic Field

Rectangular current carrying coil in a uniform magnetic field showing forces that form a couple B F₁ F₃ θ
Rectangular coil in a uniform magnetic field

Let a rectangular coil ABCD of length l and breadth b carrying current I in anticlockwise direction be placed in a uniform magnetic field B. Suppose the plane of coil makes an angle θ with magnetic field.

The forces on arms BC and DA are equal and opposite and act along the same line, so they cancel each other. The forces F₁ and F₃ form a couple and produce torque.

Force on one side: F = IlB
Torque = magnitude of either force × perpendicular distance
τ = IlB · b cosθ
Since A = lb:
τ = BIA cosθ

If rectangular coil has N turns:

τ = BINA cosθ
In the scan, θ is defined as the angle between the plane of the coil and magnetic field. Therefore the source writes the torque relation with cosθ.

Special Cases

When θ = 0°:
τ = BINA
The plane of coil is parallel to B and torque is maximum.
When θ = 90°:
τ = 0
The plane of coil is perpendicular to B and no torque acts on the coil.

Moving Coil Galvanometer

Moving coil galvanometer is a device which is used to detect and measure small electric current.

It works on the principle of torque produced on a rectangular current carrying coil placed inside the magnetic field.

Construction

Simplified construction of a moving coil galvanometer with permanent magnet, soft iron core, suspended coil, mirror and scale N pole S pole Soft iron core Rectangular coil Mirror Scale
Moving coil galvanometer – simplified construction

It consists of a rectangular coil having large number of turns wound on a non-metallic frame which is suspended between two poles of cylindrical shape permanent magnet. The coil is suspended by strip phosphor-bronze wire and strip is finally connected to the terminal of the galvanometer.

A soft iron cylinder is placed between the coil and the cylindrical poles. It makes the magnetic field stronger and radial such that in whatever position the coil rotates, magnetic field is always parallel to its plane. A concave mirror and scale arrangement is used to note the deflection.

Theory

When current I is passed through the coil, the coil gets deflected due to the resulting torque developed on it.

Magnetic torque on the coil:
τ = BINA

If k is restoring torque per unit twist and α is deflection:

Restoring torque τr = kα

At equilibrium:

kα = BINA
I = (k/BNA)α
I = Gα
where G = k/(BNA), called galvanometer constant.
I ∝ α

Current Sensitivity

It is defined as the deflection produced on the galvanometer per unit current flowing through it.
Current sensitivity = α/I = BNA/k

Voltage Sensitivity

It is defined as the deflection produced on the galvanometer per unit voltage applied to it.
Voltage sensitivity = α/V
= α/(IR)
Voltage sensitivity = BNA/(kR)

Force Between Two Parallel Current Carrying Conductors

When two parallel current carrying conductors are placed near, they exert force on each other due to magnetic field of one conductor at the position of the other conductor.

Parallel current carrying conductors showing attraction for same current direction and repulsion for opposite current direction Currents in same direction I₁ I₂ Attractive force Currents in opposite direction I₁ I₂ Repulsive force
Force between two parallel current carrying conductors

1. Currents in Same Direction

Let two infinitely long parallel conductors X and Y carry currents I₁ and I₂ respectively in same direction and be separated by distance r.

Magnetic field at conductor Y due to current I₁ in X:
B₁ = μ₀I₁/(2πr)

Force on length dl of conductor Y:

dF = I₂ dl B₁
dF = [μ₀I₁I₂/(2πr)]dl
F/l = μ₀I₁I₂/(2πr)

The force is directed towards the other conductor. Similarly equal force acts on conductor X towards conductor Y. Thus conductors attract each other when currents are in same direction.

2. Currents in Opposite Direction

When currents I₁ and I₂ are passed through two parallel conductors in opposite direction, force produced in the conductors have same magnitude but repel each other.

Why Does a Current Carrying Solenoid Try to Contract?

A solenoid tends to contract when current flows through it because the various turns of the solenoid are parallel and carry current in the same direction. Parallel wires carrying current in the same direction experience force towards each other and try to contract.
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