JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Electromagnetic Induction

A rectangular loop of length 2.5 m and width 2 m is placed at

6 0^{\circ}

to a magnetic field of 4 T . The loop is removed from the field in 10 sec . The average emf induced in the loop during this time is

(A)-2 V
(B)+2 V
(C)+1 V✓ Correct
(D)-1 V

Answer: (C)

Solution

Average emf

= \frac{∣ΔΦ∣}{Δ t}

. Initial flux with the loop at

6 0^{\circ}

to the field:

Φ_{i} = B A cos 6 0^{\circ} = 4 \times (2.5 \times 2) \times 0.5 = 10 Wb

The loop is removed, so

Φ_{f} = 0

. Over

10

∣ ε ∣ = \frac{10 - 0}{10} = 1 V

and by Lenz's law the sign works out to

+ 1

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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