A square loop of side $2.0 \mathrm{~cm}$ is placed inside a long solenoid that has 50 turns per centimetre and carries a sinusoidally varying current of amplitude $2.5 \mathrm{~A}$ 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 $x \times 10^{-4} \mathrm{~V}$. The value of $x$ is __________.

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

<p>In this problem, a square loop is inside a long solenoid, and there&#39;s a varying current flowing through the solenoid. Because the current is changing, it induces a changing magnetic field inside the solenoid. </p> <p>According to Faraday&#39;s law of electromagnetic induction, a changing magnetic field will induce an electromotive force (emf) in a loop placed in that field. In this case, the loop is the square loop inside the solenoid.</p> <p>The formula used here is based on Faraday&#39;s law, which states that the induced emf in a loop is equal to the rate of change of magnetic flux through the loop. This is given by:</p> <p>$ \text{emf} = -\frac{d \Phi}{dt} $</p> <p>where $\Phi$ is the magnetic flux. </p> <p>The magnetic field inside a solenoid is given by $B = \mu_0 n I$, where $\mu_0$ is the permeability of free space, $n$ is the number of turns per unit length in the solenoid, and $I$ is the current through the solenoid.</p> <p>The magnetic flux through the square loop is then given by $\Phi = B \cdot A = \mu_0 n I A$, where $A$ is the area of the loop. </p> <p>When the current is sinusoidal, i.e., $I(t) = I_0 \sin(\omega t)$, its derivative with respect to time is $dI/dt = I_0 \omega \cos(\omega t)$, where $\omega$ is the angular frequency.</p> <p>Hence, the rate of change of flux becomes:</p> <p>$ \frac{d \Phi}{dt} = \mu_0 n A \frac{dI}{dt} = \mu_0 n A I_0 \omega \cos(\omega t) $</p> <p>The emf, which is equal to the negative of the rate of change of flux, will have a maximum value (the amplitude) when $\cos(\omega t) = 1$, giving:</p> <p>$ \text{Emf amplitude} = \mu_0 n A I_0 \omega$ <br/><br/>$ = 4\pi \times 10^{-7} \, \text{T m/A} \times \left(\frac{50}{10^{-2}}\right) \, \text{turns/m} \times (2 \times 10^{-2} \, \text{m})^2 \times 2.5 \, \text{A} \times 700 \, \text{rad/s} $</p> <p>which simplifies to:</p> <p>$ \text{Emf amplitude} = 44 \times 10^{-4} \, \text{V} $</p> <p>So, the value of $x$ in the question is $44$</p>