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 ${{dB} \over {dt}}$ = 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 know, $\phi = BA$<br><br>Also, $E = {{d\phi } \over {dt}} = {{AdB} \over {dt}}$<br><br>$E = {l^2}{{dB} \over {dt}}$<br><br>$i = {E \over R}$ <br><br>= ${{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}$ <br><br>$= {{{l^2}} \over {pl}}{{dB} \over {dt}}A$ <br><br>= ${{{{l^2}\pi {R^2}} \over {\rho l}}{{dB} \over {dt}}}$ <br><br>$\therefore$ $$i = {{30} \over 4} \times {{30} \over 4} \times {{{{10}^{ - 4}} \times 0.032 \times 4 \times {{10}^{ - 6}} \times \pi } \over {1.23 \times {{10}^{ - 8}} \times 30 \times {{10}^{ - 2}} \times {{10}^3}}}$$<br><br>$\Rightarrow$ $i = {{240 \times \pi \times {{10}^{ - 10}}} \over {1.23 \times {{10}^{ - 7}}}}$<br><br>$\Rightarrow$ $i = {{240 \times 3.14 \times {{10}^{ - 3}}} \over {1.23}}$<br><br>$= {{753.6} \over {1.23}} \times {10^{ - 3}}$<br><br>$\Rightarrow$ $i = 612.68 \times {10^{ - 3}} = 0.61A$