Electromagnetic Induction

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Exercise 6.1Faraday's Laws and Electromagnetic Induction

Predict the direction of induced current in the situations described by the following figures: (a) A coil is placed in a changing magnetic field perpendicular to its plane (field increasing upward). (b) A rectangular loop moves to the right in a uniform magnetic field directed into the page.

Solution

(a) When field increases upward through the coil: • By Lenz's law, the induced current must create a field opposing the increase • To oppose upward increasing flux, induced current must create downward field inside the coil • Using right-hand rule: looking from above, the induced current flows clockwise (qrpq) (b) When rectangular loop moves right in a uniform field into the page: • Since the field is uniform, flux through the loop does not change as it moves • Therefore, induced emf = 0 and induced current = 0 • (If the loop were moving into or out of a region with field, current would be induced.)

Use Lenz's law to determine the direction of induced current in the following situations: (a) A magnet is quickly moved in the direction indicated in the figure towards a coil. (b) A current-carrying coil is moved towards a neighbouring coil.

Solution

(a) When N pole of magnet approaches the coil: • Flux through the coil increases (into the face facing the magnet) • By Lenz's law, induced current must oppose this increase • The face of coil facing the magnet must act as N pole (to repel the approaching magnet) • Looking from the magnet's side, the induced current flows anticlockwise (b) When current-carrying coil is moved toward the neighbouring coil: • As it approaches, flux through the neighbouring coil increases • By Lenz's law, induced current in the neighbouring coil opposes the increasing flux • The induced current flows in a direction to create a magnetic moment opposing that of the approaching coil • The two coils repel each other

A long solenoid with 15 turns per cm has a small loop of area 2.0 cm² placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing?

Solution

n = 15 turns/cm = 1500 turns/m, A = 2.0 cm² = 2.0 × 10⁻⁴ m² dI/dt = (4.0 – 2.0)/0.1 = 20 A/s Magnetic field inside solenoid: B = μ₀nI dB/dt = μ₀n(dI/dt) = 4π × 10⁻⁷ × 1500 × 20 = 4π × 10⁻⁷ × 30000 = 12π × 10⁻³ = 37.7 × 10⁻³ T/s Induced emf: ε = A × dB/dt = 2.0 × 10⁻⁴ × 37.7 × 10⁻³ = 75.4 × 10⁻⁷ ≈ 7.54 × 10⁻⁶ V ≈ 7.54 μV

A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm/s in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case?

Solution

B = 0.3 T, v = 1 cm/s = 0.01 m/s Long side = 8 cm = 0.08 m, Short side = 2 cm = 0.02 m (a) Moving out normal to the longer side (8 cm side is the leading/trailing edge): The 2 cm side cuts field lines. ε = Bvl = B × v × (shorter side) = 0.3 × 0.01 × 0.02 = 6 × 10⁻⁵ V Distance to travel = width = 8 cm = 0.08 m Time = d/v = 0.08/0.01 = 8 s (b) Moving out normal to the shorter side (2 cm side is the leading/trailing edge): The 8 cm side cuts field lines. ε = B × v × (longer side) = 0.3 × 0.01 × 0.08 = 2.4 × 10⁻⁴ V Distance to travel = 2 cm = 0.02 m Time = 0.02/0.01 = 2 s

A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad/s about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring.

Solution

l = 1.0 m, ω = 400 rad/s, B = 0.5 T For a rod rotating about one end in a uniform magnetic field perpendicular to the plane of rotation: EMF = ½Bωl² = ½ × 0.5 × 400 × (1.0)² = ½ × 0.5 × 400 = 100 V The emf developed between the centre and the ring is 100 V.

Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V is induced, give an estimate of the self-inductance of the circuit.

Solution

ΔI = 5.0 – 0 = 5.0 A, Δt = 0.1 s, ε = 200 V Using: ε = L(ΔI/Δt) 200 = L × (5.0/0.1) 200 = L × 50 L = 200/50 = 4 H Self-inductance = 4 H

A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil?

Solution

M = 1.5 H, ΔI = 20 – 0 = 20 A Change in flux linkage: ΔΦ = MΔI = 1.5 × 20 = 30 Wb The change in flux linkage is 30 Wb. Note: Average induced emf = MΔI/Δt = 1.5 × 20/0.5 = 60 V

A jet plane is travelling towards west at a speed of 1800 km/h. What is the voltage difference developed between the ends of the wing having a span of 25 m, if the Earth's magnetic field at the location has a magnitude of 5 × 10⁻⁴ T and the dip angle is 30°.

Solution

v = 1800 km/h = 500 m/s, l = 25 m, B = 5 × 10⁻⁴ T, dip = 30° The vertical component of Earth's field is responsible for the emf across the wings: Bᵥ = B sin(dip angle) = 5 × 10⁻⁴ × sin30° = 5 × 10⁻⁴ × 0.5 = 2.5 × 10⁻⁴ T EMF across wings: ε = Bᵥ × v × l = 2.5 × 10⁻⁴ × 500 × 25 = 2.5 × 10⁻⁴ × 12500 = 3.125 V ≈ 3.1 V

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Electromagnetic Induction

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