Question

A copper ring has a radius of 4 cm and resistance of 10 mΩ. A magnetic field is applied over the ring perpendicular to its plane. If the magnetic field increases from 0.2 T to 0.4 T in a time interval of 5 × 10^–3 S, what is the current in the ring during this interval?

Answer 1

dear student
$$\begin{gathered} i=\frac{e}{R}=\frac{d\varphi }{dt\times R}=\frac{dB\times S}{dt\times R}(onlymagnitude) \\ i=\frac{(0.4-0.2)\times 3.14\times 16\times {10}^{-4}}{5\times {10}^{-3}\times 10\times {10}^{-3}}=2.0096\times 10=20.096A \\ Regards \end{gathered}$$