Question

The magnetic field of a cylindrical magnet that has a pole face radius $2.8cm$ can be varied sinusoidally between the minimum value $16.8T$ and the maximum value $17.2T$ at a frequency of $50/\pi$ Hz. Cross section of the magnetic field created by the magnet is shown in figure. At a radial distance of $2cm$ from the axis, find the approximate value of the amplitude of the electric field (in units of $\times {10}^{-2}mN{C}^{-1}$) induced by the magnetic field variation.

Answer 1

Change in magnetic field $\Delta B=17.2-16.8=0.4T$

Time period $t=\frac{1}{\nu }$

Thus $\frac{\Delta B}{t}=0.4\times \frac{50}{\pi }=\frac{20}{\pi }T/s$

Using $\oint \overrightarrow{E}.\overrightarrow{dl}=\overrightarrow{A}.\frac{d\overrightarrow{B}}{dt}$

For $r<R$ $E\times 2\pi r=\pi r^2\times \frac{\Delta B}{t}$

$⟹E_r=\frac{r}{2}\frac{\Delta B}{t}$

Now at $r=2cm=2\times {10}^{-2}m$

Electric field at $r=0.02m$ $E=\frac{0.02}{2}\times \frac{20}{\pi }=6.37\times {10}^{-2}mN{C}^{-1}$