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) 35 (b) 70 (c) 7 (d) 140

Answer: (b). e = NA(Δ B)/(Δ t) with A = π r² = 22/7(0.01)² = 22/7 × 10⁻⁴ m ². (Δ B)/(Δ t) = (5000 - 3000)/(2) = 1000 T/s 22 = N × 22/7 × 10⁻⁴ × 1000 ⇒ N = 70 So the answer is (b) 70.

A coil is placed perpendicular to a magnetic field of 5000 T. When… | JEE Main 2024 (31 Jan, Shift 1)

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Magnetic Flux & Faraday's Law

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

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Answer: (b)

$e = NA\frac{\Delta B}{\Delta t}$ with $A = \pi r^2 = \frac{22}{7}(0.01)^2 = \frac{22}{7} \times 10^{-4}$ m $^2$ .
$\frac{\Delta B}{\Delta t} = \frac{5000 - 3000}{2} = 1000$ T/s
$22 = N \times \frac{22}{7} \times 10^{-4} \times 1000 \Rightarrow N = 70$
So the answer is (b) 70.

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