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 240 mN$C^{-1}$

$\int E.dt=-A.\frac{dB}{dt}$

$E\left(2\pi r\right)=-17\left[\frac{d}{dt}\right\{17+0.2\sin (\omega t+\varphi )\left\}\right]$

$\therefore E=\frac{-r}{2}\left(0.2\right)\omega \cos (\omega t+\varphi )$

$\therefore$ magnitude of amplitude $=\frac{r}{2}\left(0.2\right)=240mN/C$