Question 85 of 153

A square loop of side $2.0cm$ is placed inside a long solenoid that has $50$ turns per centimetre and carries a sinusoidally varying current of amplitude $2.5A$ and angular frequency $700rad{s}^{-1}$. The central axes of the loop and solenoid coincide. The amplitude of the emf induced in the loop is $x\times {10}^{-4}V$. The value of $x$ is ___________
$($ Take, $\left.\pi =\dfrac{22}{7}\right)$

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Answer: 44

It is given that the current is varying sinusoidally. The current can be written as $I={I}_{0}\sin \omega t$.
The magnetic field through the solenoid $B={\mu }_{0}nI$
The flux through the square is $\phi ={\mu }_{0}nIA$
The emf is
$\epsilon ={\mu }_{0}nA\times \dfrac{d({I}_{0}\sin \omega t)}{dt} \Rightarrow \epsilon ={\mu }_{0}nA{I}_{0}\omega \cos \omega t$
The amplitude of the emf is,
$\epsilon ={\mu }_{0}nA{I}_{0}\omega =4\pi \times {10}^{-7}\times \dfrac{50}{{10}^{-2}}\times 4\times {10}^{-4}\times 2.5\times 700 \Rightarrow \epsilon =44\times {10}^{-4}V$