Physics - EMI/AC Question with Solution | TestHub

As the loop is pulled out of the magnetic field, a motional EMF is induced across the side of the loop that is cutting the magnetic field lines. The magnitude of this EMF is $\text{EMF}=BLv$. Given $B=1.5\text{ T}$, $L=10\text{ cm}=0.1\text{ m}$, and $v=2\text{ m/s}$.$\text{EMF}=(1.5\text{ T})(0.1\text{ m})(2\text{ m/s})=0.3\text{ V}$. The induced current is $I=\frac{\text{EMF}}{R}$. Given $R=0.5\Omega$.$I=\frac{0.3\text{ V}}{0.5\Omega }=0.6\text{ A}$. Converting to milliamperes, $I=0.6\times 1000\text{ mA}=600\text{ mA}$.