A magnetic field B exists in a direction perpendicular to the plane of a square loop made of metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field B changes with time at a steady rate of $0.032\, T\, s^{-1}$. The induced current in the closed loop is

(Given: Resistivity of the metal wire is $1.57 x 10^{-8}Ω m$)

0.48 A

The correct answer is Option (2) → 0.48 A

Given:

Wire diameter: $d = 4 \, \text{mm} = 0.004 \, \text{m}$

Total wire length: $L = 30 \, \text{cm} = 0.3 \, \text{m}$

Rate of change of magnetic field: $\frac{dB}{dt} = 0.032 \, \text{T/s}$

Radius of wire cross-section: $r = \frac{d}{2} = 0.002 \, \text{m}$

Cross-sectional area of wire: $A_{\text{wire}} = \pi r^2 = \pi (0.002)^2 \approx 1.256 \times 10^{-5} \, \text{m}^2$

Square loop side: $a = \frac{L}{4} = \frac{0.3}{4} = 0.075 \, \text{m}$

Loop resistance: $R = \rho \frac{L}{A_{\text{wire}}}$, assuming copper wire with $\rho = 1.6 \times 10^{-8} \, \Omega \cdot \text{m}$

$R = \frac{1.6 \times 10^{-8} \cdot 0.3}{1.256 \times 10^{-5}} \approx 3.82 \times 10^{-4} \, \Omega$

Area of square loop: $A_{\text{loop}} = a^2 = (0.075)^2 = 5.625 \times 10^{-3} \, \text{m}^2$

Induced emf: $\mathcal{E} = A_{\text{loop}} \frac{dB}{dt} = 5.625 \times 10^{-3} \cdot 0.032 \approx 1.8 \times 10^{-4} \, \text{V}$

Induced current: $I = \frac{\mathcal{E}}{R} = \frac{1.8 \times 10^{-4}}{3.82 \times 10^{-4}} \approx 0.471 \, \text{A}$

Answer: $I \approx 0.47 \, \text{A}$