A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field changes with time at a steady rate dB/dt = 0.032 Ts$$^{-1}$$. The induced current in the loop is close to (Resistivity of the metal wire is 1.23 $$\times$$ 10$$^{-8}$$ $$\Omega$$m)

We are given a metal wire of total length $$L = 30\ \text{cm} = 0.30\ \text{m}$$ which is bent to form a square loop. A uniform magnetic field $$B$$ is perpendicular to the plane of the loop and is changing steadily at the rate $$\dfrac{dB}{dt}=0.032\ {\rm T\,s^{-1}}$$. The resistivity of the metal is $$\rho = 1.23 \times 10^{-8}\ \Omega\text{m}$$ and the diameter of the wire is $$d = 4\ \text{mm} = 4 \times 10^{-3}\ \text{m}$$.

First we determine the length of one side of the square. Because the wire forms a square, its four sides have equal length, so

$$4\,\ell = L \;\;\Longrightarrow\;\; \ell = \frac{L}{4} = \frac{0.30\ \text{m}}{4} = 0.075\ \text{m} = 7.5\ \text{cm}.$$

Now we find the area $$A$$ enclosed by the square loop:

$$A = \ell^{2} = (0.075\ \text{m})^{2} = 0.005625\ \text{m}^{2}.$$

The changing magnetic field produces an emf according to Faraday's law. We first state Faraday’s law in magnitude form:

$$\mathcal{E}= \left|\frac{d\Phi}{dt}\right|,$$

where the flux $$\Phi$$ through the loop is $$\Phi = BA$$. Because the field is perpendicular to the plane, the cosine factor is unity, and we can write

$$\frac{d\Phi}{dt} = A\,\frac{dB}{dt}.$$

Substituting the given numbers,

$$\frac{d\Phi}{dt} = (0.005625\ \text{m}^{2})\,(0.032\ \text{T\,s}^{-1}) = 1.8 \times 10^{-4}\ \text{Wb\,s}^{-1} = 1.8 \times 10^{-4}\ \text{V}.$$

Thus the induced emf is

$$\mathcal{E} = 1.8 \times 10^{-4}\ \text{V}.$$

Next we calculate the electrical resistance of the square loop. The cross-sectional area $$S$$ of the wire is obtained from its radius $$r$$. The radius is half the diameter:

$$r = \frac{d}{2} = \frac{4 \times 10^{-3}\ \text{m}}{2} = 2 \times 10^{-3}\ \text{m}.$$

So

$$S = \pi r^{2} = \pi\,(2 \times 10^{-3}\ \text{m})^{2} = \pi\,(4 \times 10^{-6})\ \text{m}^{2} = 4\pi \times 10^{-6}\ \text{m}^{2} \approx 1.2566 \times 10^{-5}\ \text{m}^{2}.$$

The resistance $$R$$ of a uniform wire is given by the formula

$$R = \rho\,\frac{L}{S}.$$

Substituting the known values,

$$R = \bigl(1.23 \times 10^{-8}\ \Omega\text{m}\bigr) \frac{0.30\ \text{m}}{1.2566 \times 10^{-5}\ \text{m}^{2}} = \frac{3.69 \times 10^{-9}\ \Omega\text{m}}{1.2566 \times 10^{-5}\ \text{m}^{2}} \approx 2.94 \times 10^{-4}\ \Omega = 0.000294\ \Omega.$$

The induced current $$I$$ in the loop is obtained from Ohm’s law, $$I = \dfrac{\mathcal{E}}{R}$$. Hence

$$I = \frac{1.8 \times 10^{-4}\ \text{V}}{2.94 \times 10^{-4}\ \Omega} \approx 0.612\ \text{A}.$$

This value rounds to approximately $$0.61\ \text{A}$$.

Hence, the correct answer is Option B.