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CBSE Class 12 Physics · 11 questions · 26 marks
Oersted showed that a current makes a magnetic field; Faraday asked the reverse question and changed the world with the answer. This chapter shows that a changing magnetic flux, no matter how you change it, drives an emf, and that Lenz's law fixes its direction so energy is never created out of nothing. Generators, transformers, induction cooktops and the charging pad for a phone are all standing on the two equations you will meet here.
The weber, the SI unit of magnetic flux, is equivalent to:
Answer
Volt second — from Faraday's law ε = -dΦ/dt, the flux is the product of emf and time, so 1 Wb = 1 V s. It also equals 1 T m², since Φ = BA. Tesla per square metre and ampere metre² are the units of nothing here and of magnetic moment respectively.
A straight conducting rod of length 50 cm slides on frictionless rails at a steady 4.0 m/s, perpendicular to a uniform magnetic field of 0.20 T. The emf induced across its ends is:
Answer
0.40 V — the motional emf is ε = Blv = 0.20 × 0.50 × 4.0 = 0.40 V. The rod sweeps an area lv per second, so the flux cut per second is Blv, which is exactly the emf. The polarity is fixed by Fleming's right hand rule.
The north pole of a bar magnet is pushed towards a closed circular coil. Seen from the side facing the magnet, the induced current in the coil flows:
Answer
Anticlockwise, so that the near face becomes a north pole — by Lenz's law the coil opposes the approach of the magnet, and the only way to repel an approaching north pole is to present a north pole to it. Viewed from the magnet's side, a north face requires an anticlockwise current. Work must therefore be done to push the magnet in, and that work becomes the electrical energy dissipated in the coil.
A coil of self inductance 0.40 H carries a steady current of 3.0 A. The energy stored in its magnetic field is:
Answer
1.8 J — the energy stored in an inductor is U = ½LI² = ½ × 0.40 × (3.0)² = ½ × 0.40 × 9.0 = 1.8 J. This energy is not stored in the wire but in the magnetic field threading the coil, and it is returned to the circuit when the current is switched off, which is why breaking an inductive circuit causes a spark.
Assertion (A): Lenz's law is a direct consequence of the law of conservation of energy. Reason (R): If the induced current flowed in the direction that aided the change in flux, the motion producing it would be accelerated further and electrical energy would be generated without any work being done.
Answer
Both A and R are true and R is the correct explanation of A — because the induced effects oppose the change, an external agent must do work to keep the flux changing, and that work reappears as the electrical energy dissipated in the circuit. The alternative described in R would be a perpetual motion machine, so the opposing sense demanded by Lenz's law is exactly what conservation of energy requires.
A conducting rod of length l slides with constant velocity v along a U-shaped conducting frame of total resistance R placed in a uniform magnetic field B perpendicular to its plane. Derive an expression for the induced emf and show that the mechanical power supplied equals the electrical power dissipated.
Answer
Let the rod be at distance x from the closed end of the frame, so the area of the circuit is A = lx and the flux is Φ = Blx. As the rod slides, x changes, so by Faraday's law the magnitude of the induced emf is ε = dΦ/dt = Bl (dx/dt) = Blv. The induced current is I = ε/R = Blv/R. This current, flowing in the field, experiences a force on the rod of magnitude F = BIl = B²l²v/R, directed so as to oppose the motion (Lenz's law). To keep the rod moving at constant v, an external agent must apply an equal forward force, supplying mechanical power P(mech) = Fv = B²l²v²/R. The electrical power dissipated as heat is P(elec) = I²R = (Blv/R)² R = B²l²v²/R. The two are identical, so all the mechanical work done against the magnetic force reappears as heat, confirming conservation of energy.
A flat coil of 200 turns and area 0.050 m² lies with its plane perpendicular to a magnetic field, which is increased steadily from 0.10 T to 0.40 T in 0.20 s. Find the induced emf, and the induced current if the coil has a total resistance of 30 Ω.
Answer
The field is perpendicular to the plane, so θ = 0° and Φ = BA. Change in flux per turn, ΔΦ = A ΔB = 0.050 × (0.40 - 0.10) = 0.050 × 0.30 = 1.5×10⁻² Wb. Induced emf, ε = N ΔΦ/Δt = 200 × 1.5×10⁻²/0.20 ε = 3.0/0.20 = 15 V. Induced current, I = ε/R = 15/30 = 0.50 A. By Lenz's law this current circulates so as to set up a field opposing the increase.
What are eddy currents? Give one useful application and one undesirable effect, and state how the undesirable effect is reduced.
Answer
Eddy currents are the closed loops of induced current that circulate within the body of a bulk conductor when the magnetic flux through it changes. They obey Lenz's law and therefore always oppose the change that creates them. Useful application: electromagnetic braking in trains, where a metal disc or rail moving through a field develops eddy currents whose opposing force brings the vehicle to a smooth, contactless stop. Induction furnaces and induction cooktops use the same effect to heat metal. Undesirable effect: in the iron cores of transformers and motors, eddy currents dissipate energy as heat and lower the efficiency. This is reduced by building the core from thin sheets (laminations) insulated by varnish, which confines the current loops to small cross-sections and greatly raises the resistance of the eddy paths.
(a) Define self inductance and derive an expression for the self inductance of a long air-cored solenoid of length l, area of cross-section A and N turns. (b) Derive the expression for the energy stored in an inductor carrying current I. (c) A solenoid 50 cm long, of cross-sectional area 8.0 cm², is wound with 1200 turns. Find its self inductance and the energy stored when it carries 2.5 A.
Answer
(a) Self inductance L of a coil is the flux linkage produced per unit current in the coil itself, L = NΦ/I; equivalently it is the emf induced in the coil per unit rate of change of its own current, ε = -L dI/dt. Its SI unit is the henry. Derivation: For a long solenoid with n = N/l turns per unit length carrying current I, the field inside is uniform: B = μ₀nI = μ₀NI/l. Flux through one turn: Φ = BA = μ₀NIA/l. Total flux linkage through N turns: NΦ = μ₀N²IA/l. Hence L = NΦ/I = μ₀N²A/l = μ₀n²Al. So L depends only on the geometry of the coil and the medium, not on the current. Filling the core with a material of relative permeability μ_r multiplies L by μ_r. (b) When the current in an inductor is i and is changing, the back emf is ε = L di/dt, so the source must do work at the rate dW/dt = εi = Li (di/dt). Work done in raising the current from 0 to I: W = ∫Li di from 0 to I = ½LI². Since no energy is dissipated in an ideal inductor, this work is stored in the magnetic field: U = ½LI². (c) N = 1200, l = 0.50 m, A = 8.0 cm² = 8.0×10⁻⁴ m². N² = 1.44×10⁶. L = μ₀N²A/l = (4π×10⁻⁷ × 1.44×10⁶ × 8.0×10⁻⁴)/0.50 Numerator: 1.2566×10⁻⁶ × 1.44×10⁶ = 1.8096; then 1.8096 × 8.0×10⁻⁴ = 1.448×10⁻³. L = 1.448×10⁻³/0.50 = 2.9×10⁻³ H = 2.9 mH. Energy: U = ½LI² = ½ × 2.9×10⁻³ × (2.5)² = ½ × 2.9×10⁻³ × 6.25 U = 9.1×10⁻³ J ≈ 9.1 mJ.
(a) Describe the principle and working of an AC generator and derive an expression for the instantaneous emf produced. (b) A rectangular coil of 150 turns and area 0.12 m² is rotated at 50 revolutions per second about an axis perpendicular to a uniform field of 0.050 T. Calculate the peak emf and the emf at the instant the plane of the coil is parallel to the field.
Answer
(a) Principle: an AC generator works on electromagnetic induction. When a coil is rotated in a magnetic field, the flux linked with it changes continuously, so an emf is induced in accordance with Faraday's law. Construction and working in brief: a rectangular armature coil of N turns and area A is mounted on an axle between the poles of a strong magnet, and its ends are joined to two slip rings pressed against carbon brushes. As the coil is rotated at a steady angular frequency ω, the angle between the normal to the coil and the field changes uniformly. Each half rotation reverses the sense of the flux change, so the current through the external circuit reverses direction every half cycle, giving alternating current. Derivation: Let the normal to the coil make an angle θ = ωt with the field B at time t. The flux through one turn is Φ = BA cos ωt, so the total flux linkage is NΦ = NBA cos ωt. By Faraday's law, ε = -d(NΦ)/dt = -NBA d(cos ωt)/dt ε = NBAω sin ωt. The emf is therefore sinusoidal with peak value ε₀ = NBAω, so ε = ε₀ sin ωt. The emf is maximum when the plane of the coil is parallel to the field (the coil is then cutting field lines fastest) and zero when the plane is perpendicular to the field. (b) N = 150, A = 0.12 m², B = 0.050 T, frequency ν = 50 s⁻¹. ω = 2πν = 2 × 3.14 × 50 = 314 rad/s. ε₀ = NBAω = 150 × 0.050 × 0.12 × 314 150 × 0.050 = 7.5; 7.5 × 0.12 = 0.90; 0.90 × 314 = 282.6. Peak emf ε₀ ≈ 283 V. When the plane of the coil is parallel to the field, the normal is perpendicular to B, so ωt = 90° and sin ωt = 1. The emf is then at its maximum value, 283 V.
Read the passage and answer the questions that follow: An induction cooktop hides a flat spiral coil beneath a ceramic plate. High-frequency alternating current in the coil produces a rapidly changing magnetic field, which drives circulating eddy currents in the steel base of a pan placed on top; the resistance of the steel then turns that current into heat. The ceramic plate itself, and glass or clay vessels, stay comparatively cool. A student models the pan base as a single circular loop of radius 9.0 cm and resistance 4.0×10⁻³ Ω, in which the perpendicular field rises steadily from 0 to 0.020 T in 2.5×10⁻³ s. (a) Calculate the change in magnetic flux through the modelled loop. (b) Calculate the emf induced in it during this interval. (c) Find the induced current and the rate at which heat is produced in the loop. (d) Explain why a glass vessel placed on the same cooktop does not heat up, and state the direction of the induced current relative to the coil current.
Answer
(a) Area of the loop, A = πr² = 3.14 × (0.090)² = 3.14 × 8.1×10⁻³ = 2.54×10⁻² m². ΔΦ = A ΔB = 2.54×10⁻² × 0.020 = 5.09×10⁻⁴ Wb. (b) ε = ΔΦ/Δt = 5.09×10⁻⁴/2.5×10⁻³ = 0.204 V ≈ 0.20 V. (c) Induced current, I = ε/R = 0.204/4.0×10⁻³ = 51 A. Rate of heat production, P = I²R = (51)² × 4.0×10⁻³ = 2601 × 4.0×10⁻³ = 10.4 W. The very low resistance of the metal base is exactly what allows such a large current, and hence rapid heating, from a modest emf. (d) Glass is an insulator, so although the changing flux does induce an emf in it, there are no free charge carriers to form a current loop. With no induced current there is no I²R heating, and the vessel stays cool. By Lenz's law the eddy current in a metal base circulates in the sense opposite to the growing coil current, so that its own magnetic field opposes the increase in flux through the base.
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