Chapter 6: Electromagnetic Induction
Key Formulas:
• Faraday's Law: ε = −dΦ/dt
• Induced EMF in a rod: ε = Blv
• Self-inductance: ε = −L(dI/dt)
• Mutual inductance: ε₁ = −M(dI₂/dt)
• Energy in inductor: U = ½LI²
Q1. A circular coil of area 100 cm² and 50 turns is placed perpendicular to a magnetic field of 0.2 T. The field changes at 0.01 T/s. Find the induced emf.
Solution
ε = NA(dB/dt)
= 50 × 100 × 10−4 × 0.01
= 50 × 10−2 × 0.01
= 5 × 10−3 V = 5 mV
Q2. A conducting rod of length 20 cm rotates about one end with angular velocity 10 rad/s in a uniform magnetic field of 0.3 T perpendicular to the plane of rotation. Find the induced emf.
Solution
ε = ½Bωl²
= ½(0.3)(10)(0.2)²
= ½(0.3)(10)(0.04)
= 0.06 V = 60 mV
Q3. State and explain Lenz's law. Give one application.
Solution
Lenz's Law: The direction of induced current is such that it opposes the change in magnetic flux that produced it.
Application: Electromagnetic braking in trains. When a conducting disc rotates in a magnetic field, eddy currents are induced. By Lenz's law, these currents oppose the motion, causing the disc to slow down. This principle is used in braking systems.
Q4. An inductor of 2H carries a steady current of 5A. How can a 200V emf be induced in it?
Solution
ε = L(dI/dt)
200 = 2 × (dI/dt)
dI/dt = 100 A/s
The current must change at 100 A/s to induce 200V.
Q5. Two coils have self-inductances 10 mH and 20 mH. If the mutual inductance between them is 5 mH, find the coupling coefficient.
Solution
k = M/√(L₁L₂)
= 5/√(10 × 20)
= 5/√200
= 5/(10√2)
= 1/(2√2) = 0.354
Q6. A coil of 200 turns and area 10 cm² is suddenly removed from a magnetic field of 0.1 T in 0.1s. Find the average induced emf.
Solution
ΔΦ = NBA = 200 × 0.1 × 10 × 10−4 = 0.02 Wb
ε = ΔΦ/Δt = 0.02/0.1 = 0.2 V
Q7. A rectangular coil of area A and N turns is rotated with uniform angular velocity ω in a uniform magnetic field B. Derive the expression for the instantaneous emf and show that the average emf over a complete cycle is zero.
Solution
Flux through coil: Φ = NBA cosωt
Instantaneous emf: ε = −dΦ/dt = NBAω sinωt
Average over complete cycle (T = 2π/ω):
εavg = (1/T) ∫ NBAω sinωt dt from 0 to T
= NBAω/(2π/ω) × [−cosωt/ω]02π/ω
= NBAω²/2π × [−1 + 1] = 0
Q8. A metallic rod of length l rotates with angular velocity ω about one end in a uniform magnetic field B perpendicular to the rod. Find the emf induced between the ends of the rod.
Solution
Consider a small element dx at distance x from the pivot.
Velocity of element: v = ωx
Induced emf in element: dε = Bv dx = Bωx dx
ε = ∫₀l Bωx dx = Bω[x²/2]₀l
ε = ½Bωl²
Q9. Explain eddy currents. How are they produced and what are their applications? How can they be minimized?
Solution
Eddy currents are circulating currents induced in bulk conductors when the magnetic flux through them changes.
Produced by: Changing magnetic field through a conductor or relative motion between conductor and magnet.
Applications: Electromagnetic braking, induction furnace, eddy current damping in galvanometers.
Minimization: Using laminated conductors (thin insulated sheets), or materials with high resistivity.
Q10. A coil of inductance 0.5 H carries a current that changes from 10 A to zero in 0.1 s. Find the average emf induced in the coil.
Solution
ε = L(dI/dt)
= 0.5 × (10/0.1)
= 0.5 × 100 = 50 V
Q11. Two concentric circular coils of radii r₁ and r₂ (r₂ << r₁) have N₁ and N₂ turns respectively. Find the mutual inductance of the pair.
Solution
Magnetic field due to coil 1 at centre:
B₁ = μ₀N₁I₁/(2r₁)
Flux through coil 2: Φ₂ = N₂B₁A₂ = N₂[μ₀N₁I₁/(2r₁)](πr₂²)
M = Φ₂/I₁ = μ₀N₁N₂πr₂²/(2r₁)
Q12. A square loop of side 10 cm and 20 turns is placed parallel to a uniform magnetic field of 0.2 T. The field is suddenly reversed in 0.01 s. Find the charge flowing through the coil if its resistance is 10 Ω.
Solution
ΔΦ = N(BA − (−BA)) = 2NBA
= 2 × 20 × 0.2 × (0.1)² = 0.08 Wb
ε = ΔΦ/Δt = 0.08/0.01 = 8 V
q = ε/R = 8/10 = 0.8 C
Q13. An inductor of 2H and a resistor of 10 Ω are connected in series across a 20V battery. At steady state, the battery is suddenly disconnected. What is the initial current through the inductor? How does the current decay?
Solution
At steady state: I₀ = V/R = 20/10 = 2 A
After disconnection, the current decays through the inductor and resistor:
I(t) = I₀e−Rt/L = 2e−10t/2 = 2e−5t A
Time constant: τ = L/R = 2/10 = 0.2 s
Q14. A circular loop of radius R is being deformed into a square. If the magnetic field B is uniform and perpendicular to the plane of the loop, find the charge that flows during the deformation. The resistance of the loop is r.
Solution
Initial flux: Φ₁ = B × πR²
Perimeter = 2πR, so side of square = 2πR/4 = πR/2
Final flux: Φ₂ = B × (πR/2)² = Bπ²R²/4
ΔΦ = BπR²(π/4 − 1)
q = ΔΦ/r = BπR²(1 − π/4)/r (magnitude)