1. Magnetic field inside a long solenoid

For a long solenoid, $B = \mu_0 n I$ where:

• $\mu_0 = 4\pi \times 10^{-7}\ \text{H m}^{-1}$
• $n =$ number of turns per unit length
• $I$ is the current

Given:

• Turns per centimetre $= 50$

So turns per metre, $n = 50 \times 100 = 5000\ \text{m}^{-1}$

The current varies sinusoidally with amplitude $I_0 = 2.5\ \text{A}$ and angular frequency $\omega = 700\ \text{rad s}^{-1}$

Hence magnetic field amplitude is $B_0 = \mu_0 n I_0$

2. Flux through the square loop

Side of square loop: $a = 2.0\ \text{cm} = 2.0 \times 10^{-2}\ \text{m}$

Area of loop: $A = a^2 = (2.0 \times 10^{-2})^2 = 4.0 \times 10^{-4}\ \text{m}^2$

Since the loop axis coincides with the solenoid axis, magnetic field is perpendicular to the plane of the loop, so flux is $\Phi = BA$

If $I = I_0 \sin \omega t$ then $B = \mu_0 n I_0 \sin \omega t$

Thus, $\Phi = \mu_0 n A I_0 \sin \omega t$

3. Induced emf

Induced emf is $e = \left|\frac{d\Phi}{dt}\right|$

So amplitude of induced emf is $e_0 = \mu_0 n A I_0 \omega$

Substitute values: $e_0 = (4\pi \times 10^{-7})(5000)(4.0 \times 10^{-4})(2.5)(700)$

Now simplify step-by-step:

$5000 \times 4.0 \times 10^{-4} = 2$

So, $e_0 = 4\pi \times 10^{-7} \times 2 \times 2.5 \times 700$

$2 \times 2.5 = 5$

Therefore, $e_0 = 4\pi \times 10^{-7} \times 5 \times 700$

$5 \times 700 = 3500$

Hence, $e_0 = 4\pi \times 3500 \times 10^{-7}$

$4 \times 3500 = 14000$

So, $e_0 = 14000\pi \times 10^{-7} = 14\pi \times 10^{-4}\ \text{V}$

Using $\pi = \frac{22}{7}$, $e_0 = 14 \times \frac{22}{7} \times 10^{-4} = 44 \times 10^{-4}\ \text{V}$

Thus, $x = 44$

4. Comparison with stored answer

Derived answer: $44$

Stored correct answer: $44$

They agree.