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Electromagnetic Induction and Alternating Currents - Faraday's Laws and Lenz's Law

Grade 12ICSEPhysics

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 normally through a given area. Mathematically, ΦB=B⃗⋅A⃗=BAcos⁡θ\Phi_B = \vec{B} \cdot \vec{A} = BA \cos \theta, where θ\theta is the angle between the magnetic field B⃗\vec{B} and the area vector A⃗\vec{A}. The SI unit is Weber (WbWb).

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Faraday's First Law: Whenever the magnetic flux linked with a closed circuit changes, an induced electromotive force (EMF) is produced 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 directly proportional to the rate of change of magnetic flux linked with the circuit. e∝dΦdte \propto \frac{d\Phi}{dt}.

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Lenz's Law: The direction of the induced current is such that it opposes the change in magnetic flux that produced it. This law is a direct consequence of the Law of Conservation of Energy.

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Motional EMF: When a conductor of length ll moves with a constant velocity vv in a uniform magnetic field BB such that B,l,vB, l, v are mutually perpendicular, the induced EMF is given by e=Blve = Blv.

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Fleming's Right Hand Rule: Stretch the thumb, forefinger, and middle finger of the right hand mutually perpendicular to each other. If the forefinger points in the direction of the magnetic field and the thumb in the direction of motion of the conductor, then the middle finger points in the direction of the induced current.

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Eddy Currents: These are circulating currents induced in bulk pieces of conductors when they are subjected to a changing magnetic flux. They produce a heating effect and oppose the motion (electromagnetic damping).

📐Formulae

ΦB=BAcos⁡θ\Phi_B = B A \cos \theta

e=−NdΦdte = -N \frac{d\Phi}{dt}

e=−NΦ2−Φ1Δte = -N \frac{\Phi_2 - \Phi_1}{\Delta t}

i=∣e∣R=NRdΦdti = \frac{|e|}{R} = \frac{N}{R} \frac{d\Phi}{dt}

q=∫i dt=NRΔΦq = \int i \, dt = \frac{N}{R} \Delta \Phi

e=Blve = Blv

💡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:

  1. Area of the loop A=(0.1 m)2=0.01 m2A = (0.1 \text{ m})^2 = 0.01 \text{ m}^2.
  2. The angle θ\theta between the magnetic field (North-East) and the area vector (Normal to East-West plane, i.e., North) is 45∘45^\circ.
  3. Initial Flux Φ1=BAcos⁡45∘=0.10×0.01×12≈7.07×10−4 Wb\Phi_1 = BA \cos 45^\circ = 0.10 \times 0.01 \times \frac{1}{\sqrt{2}} \approx 7.07 \times 10^{-4} \text{ Wb}.
  4. Final Flux Φ2=0\Phi_2 = 0 (since BB becomes zero).
  5. Induced EMF ∣e∣=∣ΔΦ∣Δt=7.07×10−4−00.70≈1.0×10−3 V|e| = \frac{|\Delta \Phi|}{\Delta t} = \frac{7.07 \times 10^{-4} - 0}{0.70} \approx 1.0 \times 10^{-3} \text{ V}.
  6. Induced Current i=eR=1.0×10−30.5=2.0×10−3 Ai = \frac{e}{R} = \frac{1.0 \times 10^{-3}}{0.5} = 2.0 \times 10^{-3} \text{ A}.

Explanation:

We use Faraday's Second Law to calculate the EMF based on the rate of change of flux, then apply Ohm's Law (V=IRV = IR) to find the current.

Problem 2:

A metallic rod of length 1 m1 \text{ m} is rotated with a frequency of 50 Hz50 \text{ Hz} with one end hinged at the center and the other end at the circumference of a circular metallic ring of radius 1 m1 \text{ m}, about an axis passing through the center and perpendicular to the plane of the ring. A constant magnetic field of 1 T1 \text{ T} parallel to the axis exists everywhere. Calculate the EMF induced between the center and the metallic ring.

Solution:

  1. Frequency f=50 Hzf = 50 \text{ Hz}, so angular velocity ω=2πf=100π rad/s\omega = 2\pi f = 100\pi \text{ rad/s}.
  2. Length l=1 ml = 1 \text{ m}, Magnetic field B=1 TB = 1 \text{ T}.
  3. The formula for EMF induced in a rotating rod is e=12Bωl2e = \frac{1}{2} B \omega l^2.
  4. e=12×1×100π×(1)2=50π Ve = \frac{1}{2} \times 1 \times 100\pi \times (1)^2 = 50\pi \text{ V}.
  5. e≈50×3.14=157 Ve \approx 50 \times 3.14 = 157 \text{ V}.

Explanation:

When a rod rotates in a magnetic field, different parts of the rod have different linear velocities. We integrate the motional EMF de=Bv dxde = Bv \, dx or use the average velocity vavg=ωl2v_{avg} = \frac{\omega l}{2} to get the result e=12Bωl2e = \frac{1}{2} B \omega l^2.

Faraday's Laws and Lenz's Law - Revision Notes & Key Formulas | ICSE Class 12 Physics