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 \(700 \mathrm{rad} \mathrm{s}^{-1}\). The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is \(\mathrm{x} \times 10^{-4} \mathrm{~V}\). The value of x is \_\_\_\_ (Take, \(\pi=\frac{22}{7}\))

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Electromagnetic Induction

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 \times 1 0^{- 4} V

. The value of x is ____
(Take,

π = \frac{22}{7}

)

Answer: 44

Solution

The magnetic field inside the solenoid is

B = μ_{0} n I

, where

n = 50

turns/cm

= 5000

turns/m. The flux through the square loop is

Φ = B \cdot A = μ_{0} n I A

, where

A = (2 \times 1 0^{- 2})^{2} = 4 \times 1 0^{- 4}

m². The induced emf is

ε = - \frac{d Φ}{d t} = - μ_{0} n A \frac{d I}{d t}

. For

I = I_{0} sin (ω t)

\frac{d I}{d t} = I_{0} ω cos (ω t)

. The amplitude of emf is

ε_{0} = μ_{0} n A I_{0} ω = 4 π \times 1 0^{- 7} \times 5000 \times 4 \times 1 0^{- 4} \times 2.5 \times 700 = 44 \times 1 0^{- 4}

V. Therefore,

x = 44

JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Electromagnetic Induction

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