Question

A long solenoid has $1000$ turns per meter and its radius is $R=2.5\text{cm}$. Current in the solenoid increases at a rate of $\frac{200}{\pi }\text{A/s}$. Find the magnitude of induced electric field at a point outside the solenoid at a distance $4\text{cm}$ from its axis.

• A. $3\times {10}^{-4}\text{V/m}$
• B. $6.25\times {10}^{-4}\text{V/m}$
• C. $4.5\times {10}^{-4}\text{V/m}$
• D. $9.25\times {10}^{-4}\text{V/m}$

Answer 1

The correct option is B $6.25\times {10}^{-4}\text{V/m}$

The induced electric field is obtained from expression:
$\oint E\cdot dl=\left|\frac{d\varphi }{dt}\right|$
At point $B$ at distance $r$ from centre of solenoid (outside it),
$E\left(2\pi r\right)\cos 0^{\circ }=\frac{d}{dt}\left(BS\cos 0^{\circ }\right)...\left(i\right)$
$\Rightarrow$since induced electric field is tangetial along circumference ($\theta =0^{\circ }$), and area direction along $\overrightarrow{B}$ hence angle between $B\&S$ i.e $\theta =0^{\circ }$

$R=2.5\times {10}^{-2}\text{m}$
$E\left(2\pi r\right)=\pi R^2\frac{dB}{dt}$
$\Rightarrow E=\left(\frac{R^2}{2r}\frac{dB}{dt}\right)...\left(ii\right)$
Now, for solenoid
$B={\mu }_{0}ni$
$\frac{dB}{dt}={\mu }_{0}n\left(\frac{di}{dt}\right)...\left(iii\right)$
putting the values,
$\frac{dB}{dt}=(4\pi \times {10}^{-7}\times 1000\times \left(\frac{200}{\pi }\right)$
$\frac{dB}{dt}=8\times {10}^{-2}$
From Eq $\left(ii\right)$,
$E=\frac{(2.5\times {10}^{-2}{)}^2}{2\times 4\times {10}^{-2}}\times 8\times {10}^{-2}$
$\therefore E=6.25\times {10}^{-4}\text{V/m}$