Class 12 Physics Notes – Chapter 4: Moving Charges and Magnetism

1. Introduction to Magnetic Field

Oersted discovered that a moving charge or a current-carrying wire produces a magnetic field around it. The magnetic field is a vector quantity denoted by B.

2. Magnetic Force

Force on a Moving Charge

When a charge q moves with velocity v in a magnetic field B, it experiences a magnetic force given by:

Formula: F = q(v × B) ⇒ F = qvB sinθ

Where θ is the angle between velocity and magnetic field. The force is zero if the charge is stationary (v = 0) or moving parallel to the field (θ = 0°).

Lorentz Force

If a charge moves in a region where both electric field (E) and magnetic field (B) are present, the total force is the sum of electric and magnetic forces:

F = q(E + v × B)

3. Motion in a Magnetic Field

When a charged particle enters a uniform magnetic field perpendicularly (θ = 90°), it follows a circular path. The required centripetal force is provided by the magnetic force.

• Radius of circular path: r = mv / (qB)
• Time period: T = 2πm / (qB)
• Frequency (Cyclotron frequency): ν = qB / (2πm)

Note: The time period and frequency are independent of the speed (v) and radius (r).

4. Magnetic Force on a Current-Carrying Conductor

A straight conductor of length L carrying current I placed in a uniform magnetic field B experiences a force:

Formula: F = I(L × B) ⇒ F = ILB sinθ

5. Biot-Savart Law

This law gives the magnetic field produced due to a small current element (Idl).

dB = (μ₀ / 4π) · (I dl sinθ / r²)

Where μ₀ is the permeability of free space (μ₀ = 4π × 10^-7 T m/A).

Magnetic Field on the Axis of a Circular Current Loop

For a circular loop of radius R carrying current I, having N turns, the magnetic field at its centre is:

B = (μ₀NI) / (2R)

6. Ampere’s Circuital Law

The line integral of the magnetic field B around any closed path in vacuum is equal to μ₀ times the total current threading the closed path.

Formula: ∮ B · dl = μ₀I

7. The Solenoid

A long wire wound in the form of a closely spaced helix is called a solenoid. The magnetic field inside a long current-carrying solenoid is uniform and strong.

B = μ₀nI

(where n is the number of turns per unit length, n = N/L)

8. Force Between Two Parallel Currents

Two long parallel conductors carrying currents I₁ and I₂ separated by a distance d exert a force on each other. Force per unit length is:

F / L = (μ₀ I₁ I₂) / (2πd)

• Currents in the same direction attract each other.
• Currents in opposite directions repel each other.

(This forms the basis for defining the SI unit of current, the Ampere).

9. Torque on a Current Loop and Magnetic Dipole

A current-carrying coil placed in a magnetic field experiences a torque.

τ = NIAB sinθ (or τ = m × B)

Where m is the magnetic dipole moment (m = NIA). N = turns, I = current, A = area.

10. The Moving Coil Galvanometer

It is a device used to detect small electric currents based on the principle that a current-carrying coil in a magnetic field experiences a torque.

• Conversion to Ammeter: A galvanometer is converted into an ammeter by connecting a low resistance (shunt) in parallel.
• Conversion to Voltmeter: It is converted into a voltmeter by connecting a high resistance in series.

---

Multiple Choice Questions (MCQs)

1. The SI unit of magnetic field is:
a) Weber
b) Tesla
c) Gauss
d) Oersted
Answer: (b) Tesla

2. The magnetic force on a stationary charge in a uniform magnetic field is:
a) Maximum
b) Minimum
c) Zero
d) Dependent on charge
Answer: (c) Zero (because v = 0)

3. If a charged particle enters a magnetic field perpendicularly, its path will be:
a) Straight line
b) Circular
c) Parabolic
d) Helical
Answer: (b) Circular

4. Two parallel wires carrying current in the same direction will:
a) Attract each other
b) Repel each other
c) Have no effect on each other
d) Cancel each other’s field completely
Answer: (a) Attract each other

5. The magnetic field inside a long straight solenoid carrying current is:
a) Zero
b) Non-uniform
c) Uniform
d) Maximum at the ends
Answer: (c) Uniform

6. The formula for Ampere’s circuital law is:
a) ∮ E · dl = q / ε₀
b) ∮ B · dl = μ₀I
c) ∮ B · dA = 0
d) F = q(v × B)
Answer: (b) ∮ B · dl = μ₀I

7. A galvanometer can be converted into an ammeter by connecting:
a) Low resistance in series
b) High resistance in series
c) Low resistance in parallel (shunt)
d) High resistance in parallel
Answer: (c) Low resistance in parallel (shunt)

8. The magnetic dipole moment of a current-carrying loop is given by:
a) m = N/IA
b) m = NI/A
c) m = NIA
d) m = NA/I
Answer: (c) m = NIA

9. Cyclotron frequency of a charged particle in a magnetic field is independent of:
a) Mass
b) Charge
c) Magnetic field
d) Velocity
Answer: (d) Velocity

10. The value of μ₀ (permeability of free space) is:
a) 4π × 10^-7 T m/A
b) 9 × 10⁹ N m²/C²
c) 8.85 × 10^-12 C²/N m²
d) 4π × 10⁷ T m/A
Answer: (a) 4π × 10^-7 T m/A

---

Textbook Exercise Solutions

Q 4.1: A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?
Solution:
Given: N = 100, r = 8.0 cm = 0.08 m, I = 0.40 A.
Formula: B = (μ₀NI) / (2r)
B = (4π × 10^-7 × 100 × 0.40) / (2 × 0.08)
B = (4 × 3.14 × 10^-7 × 40) / 0.16
B = 3.14 × 10^-4 T.

Q 4.2: A long straight wire carries a current of 35 A. What is the magnitude of the field B at a point 20 cm from the wire?
Solution:
Given: I = 35 A, r = 20 cm = 0.20 m.
Formula: B = μ₀I / (2πr)
B = (4π × 10^-7 × 35) / (2π &times 0.20)
B = (2 × 10^-7 × 35) / 0.20 = 3.5 × 10^-5 T.

Q 4.3: A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire.
Solution:
Given: I = 50 A, r = 2.5 m.
B = μ₀I / (2πr) = (4π × 10^-7 × 50) / (2π × 2.5) = (2 × 10^-7 × 50) / 2.5 = 4.0 × 10^-6 T.
Using Maxwell’s right-hand rule, the direction of the magnetic field at a point East of the wire is vertically upwards.

Q 4.4: A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line?
Solution:
Given: I = 90 A, r = 1.5 m.
B = μ₀I / (2πr) = (4π × 10^-7 × 90) / (2π × 1.5) = 1.2 × 10^-5 T.
By Right-hand thumb rule, the direction of magnetic field 1.5m below the wire is towards the South.

Q 4.5: What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Solution:
Given: I = 8 A, θ = 30°, B = 0.15 T.
Force per unit length, F/L = IB sinθ
F/L = 8 × 0.15 × sin(30°) = 1.2 × 0.5 = 0.6 N/m.

Q 4.6: A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire?
Solution:
Given: L = 3.0 cm = 0.03 m, I = 10 A, B = 0.27 T, θ = 90° (perpendicular).
Force F = ILB sinθ = 10 × 0.03 × 0.27 × sin(90°)
F = 10 × 0.03 × 0.27 × 1 = 8.1 × 10^-2 N.

Q 4.7: Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A.
Solution:
Given: I₁ = 8.0 A, I₂ = 5.0 A, d = 4.0 cm = 0.04 m, L = 10 cm = 0.10 m.
Force F = (μ₀ I₁ I₂ L) / (2πd)
F = (4π × 10^-7 × 8.0 × 5.0 × 0.10) / (2π × 0.04)
F = (2 × 10^-7 × 4.0) / 0.04 = 2.0 × 10^-5 N.
Since currents are in the same direction, the force is attractive.

Q 4.8: A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A, estimate the magnitude of B inside the solenoid near its centre.
Solution:
Given: Length L = 80 cm = 0.8 m. Total turns N = 5 × 400 = 2000. Current I = 8.0 A.
Turns per unit length n = N / L = 2000 / 0.8 = 2500 turns/m.
Magnetic field B = μ₀nI = (4π × 10^-7) × 2500 × 8.0
B = 4 × 3.14 × 10^-7 × 20000 = 2.5 × 10^-2 T.

Q 4.9: A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80 T. What is the magnitude of torque experienced by the coil?
Solution:
Given: Side = 10 cm, so Area A = 0.1 × 0.1 = 0.01 m². N = 20, I = 12 A, B = 0.80 T, θ = 30°.
Torque τ = NIAB sinθ
τ = 20 × 12 × 0.01 × 0.80 × sin(30°)
τ = 2.4 × 0.80 × 0.5 = 0.96 N m.

Q 4.10: Two moving coil meters, M₁ and M₂ have the following particulars: R₁ = 10 Ω, N₁ = 30, A₁ = 3.6 × 10^-3 m², B₁ = 0.25 T. R₂ = 14 Ω, N₂ = 42, A₂ = 1.8 × 10^-3 m², B₂ = 0.50 T. (Spring constants are identical). Determine ratio of (a) current sensitivity and (b) voltage sensitivity of M₂ and M₁.
Solution:
(a) Current Sensitivity I_s = NAB / k. Ratio = (N₂A₂B₂) / (N₁A₁B₁)
Ratio = (42 × 1.8 × 10^-3 × 0.50) / (30 × 3.6 × 10^-3 × 0.25) = 37.8 / 27 = 1.4.
(b) Voltage Sensitivity V_s = I_s / R. Ratio = (I_s2 / I_s1) × (R₁ / R₂)
Ratio = 1.4 × (10 / 14) = 1.0.

Q 4.11: In a chamber, a uniform magnetic field of 6.5 G (1 G = 10^-4 T) is maintained. An electron is shot into the field with a speed of 4.8 × 10⁶ m/s normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (e = 1.6 × 10^-19 C, m_e = 9.1 × 10^-31 kg).
Solution:
The electron experiences magnetic Lorentz force perpendicular to its velocity. This force acts as centripetal force, making the path circular.
Given: B = 6.5 × 10^-4 T, v = 4.8 × 10⁶ m/s.
Radius r = mv / (eB) = (9.1 × 10^-31 × 4.8 × 10⁶) / (1.6 × 10^-19 × 6.5 × 10^-4)
r = 43.68 × 10^-25 / 10.4 × 10^-23 = 4.2 × 10^-2 m = 4.2 cm.

Q 4.12: In Exercise 4.11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.
Solution:
Frequency ν = eB / (2πm) = (1.6 × 10^-19 × 6.5 × 10^-4) / (2 × 3.14 × 9.1 × 10^-31)
ν ≈ 18.18 × 10⁶ Hz = 18 MHz.
No, the frequency does not depend on the speed of the electron. The formula ν = eB / (2πm) has no velocity (v) term in it.

Q 4.13: (a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of 60° with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning.
(b) Would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area?
Solution:
(a) Given: N = 30, r = 0.08 m, I = 6.0 A, B = 1.0 T, θ = 60°.
Area A = πr² = 3.14 × (0.08)² = 0.0201 m².
Torque τ = NIAB sinθ = 30 × 6.0 × 0.0201 × 1.0 × sin(60°)
τ = 3.618 × 0.866 = 3.133 N m. (Counter torque must be of this magnitude).
(b) No, the answer would not change because torque (τ = NIAB sinθ) depends on the area (A) of the loop, not on its shape.