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Electromagnetic Induction - Faraday’s and Lenz’s Laws

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Magnetic Flux (ΦB\Phi_B): It is defined as the total number of magnetic field lines passing through a given area. It is a scalar quantity measured in Webers (WbWb). ΦB=B⃗⋅A⃗=BAcos⁡θ\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta.

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Faraday's First Law: Whenever there is a change in the magnetic flux linked with a circuit, an electromotive force (emf) is induced in the circuit, which lasts as long as the change in flux continues.

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Faraday's Second Law: The magnitude of the induced emf is equal to the time rate of change of magnetic flux through the circuit. ε∝dΦBdt\varepsilon \propto \frac{d\Phi_B}{dt}.

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Lenz's Law: The direction of induced current is such that it opposes the change in magnetic flux that produced it. It is a manifestation of the law of conservation of energy.

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Motional EMF: When a conductor of length ll moves with a velocity vv in a uniform magnetic field BB such that B,l,B, l, and vv are mutually perpendicular, the induced emf is ε=Blv\varepsilon = Blv.

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Self-Induction: The property of a coil by virtue of which it opposes any change in the strength of current flowing through it by inducing an emf in itself. The constant of proportionality is Self-Inductance (LL).

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Mutual Induction: The phenomenon of inducing an emf in a secondary coil due to a change of current in the primary coil. The constant is Mutual Inductance (MM).

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Eddy Currents: When a bulk conductor is placed in a changing magnetic field, induced currents circulate within the body of the conductor. These are called Eddy currents and result in heating and electromagnetic damping.

📐Formulae

ΦB=BAcos⁡θ\Phi_B = B A \cos \theta

ε=−NdΦBdt\varepsilon = -N \frac{d\Phi_B}{dt}

ε=Blv\varepsilon = B l v

ε=12Bωl2\varepsilon = \frac{1}{2} B \omega l^2

Φ=LI  ⟹  ε=−LdIdt\Phi = L I \implies \varepsilon = -L \frac{dI}{dt}

Φ2=MI1  ⟹  ε2=−MdI1dt\Phi_2 = M I_1 \implies \varepsilon_2 = -M \frac{dI_1}{dt}

L=μ0N2AlL = \frac{\mu_0 N^2 A}{l}

U=12LI2U = \frac{1}{2} L I^2

💡Examples

Problem 1:

A square loop of side 10 cm10 \text{ cm} and resistance 0.5 Ω0.5 \, \Omega is placed vertically in the east-west plane. A uniform magnetic field of 0.10 T0.10 \text{ T} is set up across the plane in the north-east direction. The magnetic field is decreased to zero in 0.70 s0.70 \text{ s} at a steady rate. Determine the magnitude of induced emf and current during this time interval.

Solution:

Area A=(0.1)2=10−2 m2A = (0.1)^2 = 10^{-2} \text{ m}^2. The angle θ\theta between the normal to the loop (North or South) and the magnetic field (North-East) is 45∘45^\circ. Initial flux Φ1=BAcos⁡45∘=0.1×10−2×12≈0.707×10−3 Wb\Phi_1 = BA \cos 45^\circ = 0.1 \times 10^{-2} \times \frac{1}{\sqrt{2}} \approx 0.707 \times 10^{-3} \text{ Wb}. Final flux Φ2=0\Phi_2 = 0. Δt=0.70 s\Delta t = 0.70 \text{ s}. Magnitude of emf ∣ε∣=∣ΔΦΔt∣=0.707×10−30.7≈1.0×10−3 V|\varepsilon| = |\frac{\Delta \Phi}{\Delta t}| = \frac{0.707 \times 10^{-3}}{0.7} \approx 1.0 \times 10^{-3} \text{ V}. Current I=εR=10−30.5=2×10−3 A=2 mAI = \frac{\varepsilon}{R} = \frac{10^{-3}}{0.5} = 2 \times 10^{-3} \text{ A} = 2 \text{ mA}.

Explanation:

Using Faraday's law, we calculate the change in flux over time. The angle is 45∘45^\circ because the field is North-East while the loop area vector is North (perpendicular to east-west).

Problem 2:

A wheel with 1010 metallic spokes each 0.5 m0.5 \text{ m} long is rotated with a speed of 120 rev/min120 \text{ rev/min} in a plane normal to the horizontal component of earth's magnetic field BH=0.4 GB_H = 0.4 \text{ G} at a place. What is the induced emf between the axle and the rim of the wheel? (1 G=10−4 T1 \text{ G} = 10^{-4} \text{ T})

Solution:

l=0.5 ml = 0.5 \text{ m}, f=120/60=2 rev/sf = 120 / 60 = 2 \text{ rev/s}, B=0.4×10−4 TB = 0.4 \times 10^{-4} \text{ T}. Angular velocity ω=2πf=4π rad/s\omega = 2 \pi f = 4 \pi \text{ rad/s}. The induced emf is ε=12Bωl2=12×(0.4×10−4)×4π×(0.5)2\varepsilon = \frac{1}{2} B \omega l^2 = \frac{1}{2} \times (0.4 \times 10^{-4}) \times 4 \pi \times (0.5)^2. ε=0.2×10−4×4π×0.25=0.2π×10−4≈6.28×10−5 V\varepsilon = 0.2 \times 10^{-4} \times 4 \pi \times 0.25 = 0.2 \pi \times 10^{-4} \approx 6.28 \times 10^{-5} \text{ V}.

Explanation:

The formula for a rotating rod is used. Note that the number of spokes does not affect the potential difference between the axle and the rim because they are all in parallel.