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 \( 0.032\,\mathrm{T\,s^{-1}} \). The induced current in the closed loop is
(Given: Resistivity of the metal wire is \( 1.57\times10^{-8}\,\Omega\,\mathrm{m} \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 0.48\,\mathrm{A} \)
By Faraday's law a changing field through the loop induces an emf \( \varepsilon = A\,(dB/dt) \), and the current is this emf divided by the resistance of the wire, \( R = \rho L/A_{\mathrm{wire}} \).
The 30 cm of wire is bent into a square, so each side is \( 30/4 = 7.5\,\mathrm{cm} \).
\( \displaystyle A_{\mathrm{loop}} = (7.5\times10^{-2})^{2} = 5.625\times10^{-3}\,\mathrm{m^{2}} \)
\( \displaystyle \varepsilon = A_{\mathrm{loop}}\,\frac{dB}{dt} = 5.625\times10^{-3} \times 0.032 = 1.8\times10^{-4}\,\mathrm{V} \)
The wire has a circular cross-section of diameter \( 4\,\mathrm{mm} \), so its radius is \( 2\times10^{-3}\,\mathrm{m} \):
\( \displaystyle A_{\mathrm{wire}} = \pi r^{2} = \pi\,(2\times10^{-3})^{2} = 1.257\times10^{-5}\,\mathrm{m^{2}} \)
\( \displaystyle R = \frac{\rho L}{A_{\mathrm{wire}}} = \frac{1.57\times10^{-8} \times 0.30}{1.257\times10^{-5}} = 3.75\times10^{-4}\,\Omega \)
\( \displaystyle I = \frac{\varepsilon}{R} = \frac{1.8\times10^{-4}}{3.75\times10^{-4}} = 0.48\,\mathrm{A} \)