Question

A uniform magnetic field B exists in a direction perpendicular to the plane of a square frame made of copper wire. The wire has a diameter of 2 mm and a total length of 40 cm. The magnetic field changes with time at a steady rate $dB/dt=0.02T{s}^{-1}$. Find the current induced in the frame. Resistivity of copper $={1.7\times 10}^{-5}Wm.$

Answer 1

$P=$ Resistivity of copper $=1.7\times {10}^{-5}{W}_m$

diameter of $=2mm$

total length $=40cm$

The magnetic field change with time at a steady rate $=\frac{dB}{dt}=0.02T/s$.

$I=\frac{e}{R}$

$=\frac{1}{R}.\left(\frac{dc}{dt}\right)$

$=\frac{A}{R}\left(\frac{dB}{dt}\right)$

$=A.\frac{dB}{dt}.\frac{a}{pl}$

$=A\left(\frac{dB}{dt}\right).\pi r^2/pl$

Substituting the value and you get the answer

$=\frac{0.02\times \pi \times {(\frac{2}{10}\frac{2}{2}\times {10}^{-3})}^2}{1.7\times {10}^{-5}\times 40\times {10}^{-2}}$

$=\frac{0.02\times 3.14\times {10}^{-6}}{1.7\times {10}^{-5}\times 40\times {10}^{-2}}$

$=9.2mA$.