Question

The magnetic field of a cylindrical magnet that has a pole-face radius $2.8 \,cm$ can be varied sinusoidally between minimum value $16.8 \,T$ and maximum value $17.2 \,T$ at a frequency of $\frac{60}{\pi}\,Hz$ Cross section of the magnetic field created by the magnet is shown. At a radial distance of $2\, cm$ from the axis, find the amplitude of the electric field (in $mN \,C^{-1})$ induced by the magnetic field variation.

• A. $240\,mN\,C^{-1}$
• B. $180\,mN\,C^{-1}$
• C. $110\,mN\,C^{-1}$
• D. $290\,mN\,C^{-1}$

Answer 1

The Correct Option is A

$\int \vec{E} \cdot\vec{dl}=-A \frac{dB}{dt}$ As $B=17+\left(0.2\right)sin \left(\omega t+\phi\right)$ $E\left(2\pi r\right)=-\pi r^{2}\left(0.2\right)\omega\,cos\left(\omega t+\phi\right)$ $E=-\frac{r}{2}\left(0.2\right)\omega\,cos\left(\omega t+\phi\right)$ Magnitude of the amplitude $=\frac{r}{2}\left(0.2\right)\omega=240\,mN\,C^{-1}$