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PhysicsElectromagnetic Induction
A rectangular wire loop of sides 8 cm and 3 cm with a small cut, is moving out of a region of uniform magnetic field of magnitude 0·3 T directed normal to the plane of the loop. The velocity of the loop is 2 cm s⁻¹, in a direction normal to the shorter side of the loop. The emf developed across the cut, if the velocity of the loop is in the direction normal to the shorter side of the loop, will be :
Options
1
4·8 × 10⁻⁴ volt
2
1·2 × 10⁻⁴ volt
3
1·3 × 10⁻⁴ volt
4
1·8 × 10⁻⁴ volt
Correct Answer
Option 4 : 1·8 × 10⁻⁴ V
Step-by-Step Solution
1

Identify the effective length:

Velocity is normal to the shorter side (3 cm). This means the loop moves perpendicular to the 3 cm side.

The side that cuts through field lines as the loop exits = the side parallel to itself in the direction of motion = the shorter side = 3 cm = 0.03 m

2

Convert units:

l = 3 cm = 0.03 m (shorter side, effective cutting length)

v = 2 cm/s = 0.02 m/s

B = 0.3 T

3

Apply motional EMF formula:

\(\varepsilon = Bvl = 0.3 \times 0.02 \times 0.03\)

\(= 0.3 \times 6 \times 10^{-4} = 1.8 \times 10^{-4} \text{ V}\) ✓

Theory: Electromagnetic Induction
1. Faraday's Laws of EMI

First Law: Whenever magnetic flux through a circuit changes, an EMF is induced in it. Second Law: The magnitude of induced EMF = rate of change of magnetic flux: ε = −dΦ/dt. The negative sign indicates Lenz's law (induced current opposes the change causing it).

\(\varepsilon = -\dfrac{d\Phi}{dt}\)

Flux: Φ = BA cosθ

2. Motional EMF

When a conductor of length l moves with velocity v perpendicular to a magnetic field B, the free electrons in the conductor experience a force F = qv × B, creating a potential difference (EMF) across the conductor:

\(\varepsilon = Bvl\) (when B, v, l all mutually perpendicular)

This is also consistent with Faraday's law: as the loop exits the field region, flux decreases at rate dΦ/dt = B × (rate of area leaving field) = B × v × l, giving ε = Bvl.

3. Which Side is the Effective Length?

This is the key conceptual point in this problem. The loop has sides 8 cm (longer) and 3 cm (shorter). The velocity is normal to (perpendicular to) the shorter side — meaning the loop moves in the direction of the longer dimension. As the loop exits the field region:

📌 The side at the boundary between field and no-field region = the shorter side (3 cm)

📌 This 3 cm side sweeps out field lines as it moves with the loop

📌 The two longer sides (8 cm) move parallel to their own length — no cutting of field lines

📌 Effective length l = 3 cm = shorter side

4. Lenz's Law

The direction of induced current is such that it opposes the cause producing it. As the loop moves out of the magnetic field, flux decreases. By Lenz's law, the induced current must try to maintain the flux — so it flows in a direction to produce magnetic field in the same direction as the original field (using right-hand rule). This creates an opposing force on the loop, requiring external force to maintain constant velocity.

5. Key EMI Formulas for NEET

📌 ε = −dΦ/dt (Faraday's law)

📌 ε = Bvl (motional EMF, B⊥v⊥l)

📌 ε = NBAω sinωt (AC generator)

📌 ε_max = NBAω (peak EMF of AC generator)

📌 Lenz's law: induced current opposes flux change

6. Self and Mutual Inductance

Self-inductance L: EMF induced in a coil due to change in its own current: ε = −L(dI/dt). Unit: Henry (H). Mutual inductance M: EMF induced in secondary coil due to change in current in primary: ε₂ = −M(dI₁/dt). Transformer works on mutual inductance: V₁/V₂ = N₁/N₂ (ideal transformer).

Frequently Asked Questions
1. Why is the effective length 3 cm and not 8 cm?
Velocity is normal to the shorter side (3 cm), meaning the loop moves perpendicular to the 3 cm side. As the loop exits the field, the 3 cm side is at the boundary and sweeps through field lines. The 8 cm sides move parallel to their own length — they don't cut field lines. So l = 3 cm.
2. What is motional EMF formula?
ε = Bvl when B, v, and l are mutually perpendicular. Here B = 0.3T (perpendicular to loop plane), v = 0.02 m/s, l = 0.03 m. ε = 0.3 × 0.02 × 0.03 = 1.8×10⁻⁴ V.
3. What does "normal to the plane of loop" mean for B?
B is perpendicular to the plane of the rectangular loop — i.e., B points straight through the loop (like sticking through the paper if the loop lies flat on paper). This maximises flux: Φ = BA (since cosθ = cos0° = 1).
4. What is Lenz's law here?
As the loop exits the field, flux through it decreases. Lenz's law: induced current flows to oppose this decrease — it flows in a direction to create magnetic field in the same direction as B (into the loop plane). This creates a force opposing the loop's motion (opposing exit from field).
5. Why does the loop have a small cut?
The cut prevents current from flowing around the loop. Without a cut, induced current would flow and there would be both an EMF source and a resistance. With the cut, it is an open circuit — we measure the open-circuit EMF (which equals the Faraday EMF = Bvl) across the gap.
6. What would the EMF be if velocity were normal to the longer side?
Then the 8 cm side would be at the boundary: ε = Bvl = 0.3 × 0.02 × 0.08 = 4.8×10⁻⁴ V. This is option 1 — the trap answer for students who use the wrong length.
7. What is the flux change rate in this problem?
dΦ/dt = B × (rate of change of area) = B × v × l = 0.3 × 0.02 × 0.03 = 1.8×10⁻⁴ Wb/s. This equals the EMF by Faraday's law (ε = dΦ/dt). Both Faraday's law and motional EMF formula give the same answer.
8. What is the self-inductance of a solenoid?
L = μ₀N²A/l = μ₀n²Al, where N = total turns, A = cross-sectional area, l = length, n = N/l = turns per unit length. Inductance is purely geometric for an ideal solenoid — depends on shape and size, not on current.
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