1. Find the side of the square loop

The total length of the wire is the perimeter of the square: $4a = 30\text{ cm} = 0.30\text{ m}$ So, $a = \frac{0.30}{4} = 0.075\text{ m}$

2. Area of the square loop

$A = a^2 = (0.075)^2 = 5.625\times 10^{-3}\text{ m}^2$

3. Induced emf

Since the magnetic field is perpendicular to the plane of the loop, $\Phi = BA$ Hence, $\mathcal{E} = \left|\frac{d\Phi}{dt}\right| = A\frac{dB}{dt}$

Given: $\frac{dB}{dt} = 0.032\text{ T s}^{-1}$ So, $\mathcal{E} = 5.625\times 10^{-3}\times 0.032 = 1.8\times 10^{-4}\text{ V}$

4. Resistance of the wire

Resistance is $R = \rho \frac{L}{A_c}$ where:

• $\rho = 1.23\times 10^{-8}\ \Omega\text{m}$
• $L = 0.30\text{ m}$
• $A_c$ is cross-sectional area of the wire

The wire diameter is $4\text{ mm}$, so radius is $r = 2\text{ mm} = 2\times 10^{-3}\text{ m}$ Cross-sectional area: $A_c = \pi r^2 = \pi(2\times 10^{-3})^2 = 4\pi\times 10^{-6}\text{ m}^2$

Thus, $R = \frac{1.23\times 10^{-8}\times 0.30}{4\pi\times 10^{-6}}$ $R \approx 2.94\times 10^{-4}\ \Omega$

5. Induced current

Using Ohm’s law, $I = \frac{\mathcal{E}}{R} = \frac{1.8\times 10^{-4}}{2.94\times 10^{-4}} \approx 0.61\text{ A}$

6. Check options

The value is closest to: $\boxed{0.61\text{ A}}$ So, Option D is correct.