JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Electromagnetic Induction

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A coil is placed perpendicular to a magnetic field of 5000 T. When the field is changed to 3000 T in 2 s, an induced emf of 22 V is produced in the coil. If the diameter of the coil is 0.02 m, then the number of turns in the coil is:

(A)7
(B)70✓ Correct
(C)35
(D)140

Answer: (B)

Solution

Induced emf:

ε = N \frac{d Φ}{d t} = N \cdot A \cdot \frac{Δ B}{Δ t}

, with

A = π r^{2}

.

Diameter

0.02 m \Rightarrow r = 0.01 m \Rightarrow A = π (0.01)^{2} = π \times 1 0^{- 4} m^{2}

Δ B = 5000 - 3000 = 2000 T

over

Δ t = 2 s

22 = N \times π \times 1 0^{- 4} \times \frac{2000}{2} = N \times 0.1 π

N = \frac{22}{0.1 π} \approx 70

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Single correct option

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