JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Electromagnetic Induction

A square loop of side 15 cm being moved towards right at a constant speed of

2 cm / s

as shown in figure. The front edge enters the 50 cm wide magnetic field at

t = 0

. The value of induced emf in the loop at

t = 10 s

will be :

(A)0.3 mV
(B)4.5 mV
(C)zero✓ Correct
(D)3 mV

Answer: (C)

Solution

The loop takes

50/2 = 25

s to fully enter the 50 cm wide field, so by

t = 10

s the leading edge has moved

2 \times 10 = 20

cm — the loop (15 cm side) is already completely inside the field (since

20 cm > 15 cm

, the trailing edge entered at

t = 7.5

s).

Once the loop is entirely inside a uniform field, the flux through it is constant, so

ε = - \frac{d ϕ}{d t} = 0

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option

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