Exercises
(6.1) Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f).
Answer
In each case we apply Lenz's law: the induced current flows so as to oppose the change in magnetic flux through the coil.
(a) The south pole of the magnet moves towards the coil, so the face of the coil near the magnet must behave as a south pole (to repel it). Induced current flows along qrpq.
(b) The magnet moves left: its S pole approaches the left coil and its N pole recedes from the right coil. The near face of the left coil becomes S and the near face of the right coil becomes S (to attract the receding N pole). Current flows along prq in the left coil and along yzx in the right coil.
(c) When the tapping key is just closed, current grows in the left loop and the flux through the right loop increases. The induced current in the right loop opposes this, flowing in the opposite sense: along yzx.
(d) Changing the rheostat setting increases the current in the right loop, so flux through the left loop increases; the induced current flows to oppose it: along zyx.
(e) When the key is just released, the current in the left coil falls to zero and the flux through the right coil decreases. The induced current tries to maintain the flux and flows along xry.
(f) The magnetic field lines of the straight current form circles around the wire, which lie in the plane of the loop. No field lines pass through the loop, so the flux is zero and stays zero: no current is induced.
(6.2) Use Lenz's law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; (b) A circular loop being deformed into a narrow straight wire.
Answer
(a) The field is directed into the page. When the irregular wire becomes a circle, its enclosed area increases, so the inward flux increases. The induced current must create an outward field, i.e. it flows anticlockwise as seen: along adcb.
(b) The field is directed out of the page. As the circle is squeezed into a narrow wire, the enclosed area and hence the outward flux decrease. The induced current tries to keep up the outward flux, so it flows anticlockwise: along a′d′c′b′.
(6.3) 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?
Answer
Given: n = 15 turns/cm = 1500 m⁻¹, A = 2.0 cm² = 2.0 × 10⁻⁴ m², ΔI = 4.0 − 2.0 = 2.0 A, Δt = 0.1 s.
Field inside the solenoid B = μ₀nI, so flux through loop Φ = μ₀nIA.
e = dΦ/dt = μ₀nA (ΔI/Δt)
⇒ e = 4π × 10⁻⁷ × 1500 × 2.0 × 10⁻⁴ × (2.0/0.1)
⇒ e ≈ 7.5 × 10⁻⁶ V
(6.4) 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⁻¹ 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?
Answer
Given: B = 0.3 T, v = 1 cm s⁻¹ = 0.01 m s⁻¹. Motional emf e = Blv, where l is the side perpendicular to the velocity (the side cutting the field lines).
(a) Velocity normal to the longer side, so l = 8 cm = 0.08 m
⇒ e = 0.3 × 0.08 × 0.01 = 2.4 × 10⁻⁴ V.
The loop must move a distance equal to the shorter side (2 cm) to leave the field
⇒ time = 2 cm ÷ 1 cm s⁻¹ = 2 s.
(b) Velocity normal to the shorter side, so l = 2 cm = 0.02 m
⇒ e = 0.3 × 0.02 × 0.01 = 0.6 × 10⁻⁴ V = 6 × 10⁻⁵ V.
Distance to leave the field = 8 cm
⇒ time = 8 cm ÷ 1 cm s⁻¹ = 8 s.
(6.5) A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s⁻¹ 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.
Answer
Given: l = 1.0 m, ω = 400 rad s⁻¹, B = 0.5 T.
An element dr at distance r moves with speed ωr, so de = Bωr dr. Integrating from 0 to l gives e = ½Bωl².
e = ½ × 0.5 × 400 × (1.0)² → e = 100 V
(6.6) A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s⁻¹, at right angles to the horizontal component of the earth's magnetic field, 0.30 × 10⁻⁴ Wb m⁻². (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential?
Answer
Given: l = 10 m, v = 5.0 m s⁻¹, horizontal field B = 0.30 × 10⁻⁴ T.
(a) e = Blv = 0.30 × 10⁻⁴ × 10 × 5.0
⇒ e = 1.5 × 10⁻³ V
(b) The horizontal field points from south to north and the velocity is vertically downward. The force on a positive charge, q(v × B), points towards the east. Hence the emf acts from west to east along the wire.
(c) Positive charges accumulate at the eastern end, so the eastern end is at the higher potential.
(6.7) Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit.
Answer
Given: ΔI = 0 − 5.0 = −5.0 A, Δt = 0.1 s, |e| = 200 V.
|e| = L |ΔI/Δt|
⇒ L = |e| Δt / |ΔI|
⇒ L = 200 × 0.1 / 5.0
⇒ L = 4 H
(6.8) 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?
Answer
Given: M = 1.5 H, ΔI = 20 A, Δt = 0.5 s.
Flux linkage with the second coil NΦ = MI, so Δ(NΦ) = M ΔI
⇒ Δ(NΦ) = 1.5 × 20
⇒ Δ(NΦ) = 30 Wb
(The time interval is not needed for the change in flux linkage; it would give the average induced emf, 30/0.5 = 60 V.)
Important Extra Questions
State Faraday's law of electromagnetic induction.
Answer
The magnitude of the emf induced in a circuit equals the time rate of change of magnetic flux linked with it: e = −N dΦ/dt for a coil of N turns. The negative sign expresses Lenz's law.
Show that Lenz's law is a consequence of the conservation of energy.
Answer
If the induced current aided the change producing it, a magnet pushed slightly towards a coil would be pulled in faster, gaining kinetic energy and producing more current with no work done—a perpetual source of energy. Since that is impossible, the induced current must oppose the change, and the work done against this opposition appears as electrical energy (Joule heat).
Define magnetic flux. Is it a scalar or a vector? Give its SI unit.
Answer
Magnetic flux through a surface of area A in a uniform field B is Φ = B·A = BA cos θ, where θ is the angle between B and the normal to the surface. It is a scalar quantity; its SI unit is the weber (Wb) = T m².
Write the expression for the self-inductance of a long solenoid and the energy stored in an inductor.
Answer
L = μᵣμ₀n²Al, where n is turns per unit length, A the cross-section area and l the length. Energy stored when current I flows: U = ½LI².
A coil of N turns and area A rotates with angular speed ω in a uniform field B. Write the induced emf and its peak value.
Answer
e = NBAω sin ωt; peak emf e₀ = NBAω, reached when the plane of the coil is parallel to the field (flux is then zero but changing fastest).