Electromagnetic Induction

JEE Main level, single correct, medium

115 of 153

Question 115 of 153

A 3030 cm long solenoid has 1010 turns per cm and area of 55 cm2^2. The current through the solenoid coil varies from 22 A to 44 A in 3.143.14 s. The e.m.f. induced in the coil is α×10−5\alpha\times 10^{-5} V. The value α\alpha is ________.

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Answer: (B) 1212

Given:
Length of the solenoid, l=30 cm=0.3 ml = 30 \text{ cm} = 0.3 \text{ m}
Number of turns per unit length, n=10 turns/cm=1000 turns/mn = 10 \text{ turns/cm} = 1000 \text{ turns/m}
Area of cross-section, A=5 cm2=5×10−4 m2A = 5 \text{ cm}^2 = 5 \times 10^{-4} \text{ m}^2
Change in current, ΔI=4 A−2 A=2 A\Delta I = 4 \text{ A} - 2 \text{ A} = 2 \text{ A}
Time interval, Δt=3.14 s\Delta t = 3.14 \text{ s}

The self-inductance of the solenoid is given by:
L=μ0n2AlL = \mu_0 n^2 A l

Substituting the given values:
L=4π×10−7×(1000)2×5×10−4×0.3L = 4\pi \times 10^{-7} \times (1000)^2 \times 5 \times 10^{-4} \times 0.3
L=4π×10−7×106×1.5×10−4L = 4\pi \times 10^{-7} \times 10^6 \times 1.5 \times 10^{-4}
L=6π×10−5 HL = 6\pi \times 10^{-5} \text{ H}

The magnitude of the induced e.m.f. is:
e=LΔIΔte = L \dfrac{\Delta I}{\Delta t}

Substituting the values of LL, ΔI\Delta I, and Δt\Delta t:
e=6π×10−5×23.14e = 6\pi \times 10^{-5} \times \dfrac{2}{3.14}

Using π≈3.14\pi \approx 3.14:
e=6×10−5×2=12×10−5 Ve = 6 \times 10^{-5} \times 2 = 12 \times 10^{-5} \text{ V}

Comparing this with e=α×10−5 Ve = \alpha \times 10^{-5} \text{ V}, we get:
α=12\alpha = 12

Answer: 1212

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