Question

The magnetic field of a cylindrical magnet that has a pole-face radius $2.8\text{cm}$ can be varied sinusoidally between minimum value of $14.8\text{T}$ and maximum value of $15.2\text{T}$ at a frequency of $50/\pi \text{Hz}$. The cross-section of magnetic field is created by the magnet is shown. At a radial distance of $2\text{cm}$ from the axis, the amplitude of electric field induced by the magnetic field variation $\left(\text{in mN/C}\right)$ is
(Integer only)

Answer 1

We know that,

$\oint \overrightarrow{E}\cdot \overrightarrow{dl}=-\frac{d{\varphi }_B}{dt}$

$\because {\varphi }_B=\overrightarrow{B}\cdot \overrightarrow{A}$

$\Rightarrow \oint \overrightarrow{E}\cdot \overrightarrow{dl}=-\overrightarrow{A}\cdot \frac{\overrightarrow{dB}}{dt}......\left(1\right)$

Let $\omega$ be the angular frequency.

Since magnetic field intensity varies from $14.8\text{T}$ to $15.2\text{T}$, so according to problem, magnetic field will be

$B=15+\left(0.2\right)\sin (\omega t+\varphi )$

$\Rightarrow \frac{dB}{dt}=\left(0.2\right)\omega \cos (\omega t+\varphi )....\left(2\right)$

Substituting (2) in (1),

$E\left(2\pi r\right)=-\pi r^2\left(0.2\right)\omega \cos (\omega t+\varphi )$

$\Rightarrow E=-\frac{r\omega }{10}\cos (\omega t+\varphi )$

The maximum amplitude of Electric field is given by,

$E_0=\frac{r\omega }{10}$

Here, $\omega =2\pi \nu =2\pi \times \frac{50}{\pi }=100\text{rad/s}$

$E_0=\left(\frac{0.02\times 100}{10}\right)$

$\Rightarrow E_0=200\times {10}^{-3}{\text{NC}}^{-1}=200{\text{mNC}}^{-1}$

Correct answer$:200$