A square loop of side 2.0 cm is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude 2.5 A and angular frequency 700 rad s⁻¹. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is x × 10⁻⁴ V. The value of x is __________. (Take, π = 22/7) Numerical answer type. Work it out, then open the answer.

Answer: 44. B = μ₀ n I with n = 50 per cm = 5000 m⁻¹; emf amplitude e₀ = μ₀ n I₀ ω A e₀ = 4π×10⁻⁷×5000×2.5×700×(0.02)² e₀ = 1.4π×10⁻³ = 1.4×22/7×10⁻³ = 4.4×10⁻³ = 44×10⁻⁴ V So x = 44.

A square loop of side 2.0 cm is placed inside a long solenoid that… | JEE Main 2023 (10 Apr, Shift 2)

Q47JEE Main 2023 · 10 APR · SHIFT 2Difficult

Mutual Inductance

A square loop of side $2.0\text{ cm}$ is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude $2.5\text{ A}$ and angular frequency $700\text{ rad s}^{-1}$ . The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is $x \times 10^{-4}\text{ V}$ . The value of $x$ is __________.

(Take, $\pi = \frac{22}{7}$ )

Numerical answer type. Work it out, then open the answer.

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Answer: 44

$B = \mu_0 n I$ with $n = 50\text{ per cm} = 5000\text{ m}^{-1}$ ; emf amplitude $e_0 = \mu_0 n I_0 \omega A$
$e_0 = 4\pi\times10^{-7}\times5000\times2.5\times700\times(0.02)^2$
$e_0 = 1.4\pi\times10^{-3} = 1.4\times\frac{22}{7}\times10^{-3} = 4.4\times10^{-3} = 44\times10^{-4}\text{ V}$
So $x = 44$ .

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