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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.

Faraday’s and Lenz’s Laws - Revision Notes & Key Formulas | CBSE Class 12 Physics