Question

The magnitude of the induced electric field inside the solenoid, at a distance $r<R$ from its long central axis is :

• A. $\frac{3{\mu }_{0}n{I}_{max}\omega }{2}r\sin \omega t$
• B. $\frac{{\mu }_{0}n{I}_{max}\omega }{2}r\cos \omega t$
• C. ${\mu }_{0}n{I}_{max}\omega r\sin \omega t$
• D. $\frac{{\mu }_{0}n{I}_{max}\omega }{2}r\sin \omega t$

Answer 1

Answer: (D)

$\frac{{\mu }_{0}n{I}_{max}\omega }{2}r\sin \omega t$

Magnetic field inside the solenoid ,

$B={\mu }_{0}nI$

Flux through the area of radius r,

$\varphi =\int B.ds={\mu }_{0}nI\left(\pi r^2\right)$

$|\oint \overrightarrow{E.}dl|=\left|\frac{d\varphi m}{dt}\right|$

$\overrightarrow{E}.\left(2\pi r\right)=\left|{\mu }_{0}n\left(\pi r^2\right)\frac{dI}{dt}\right|$

$\overrightarrow{E}=\left|\frac{{\mu }_{0}nr}{2}\frac{d\left(I_{max}\cos \omega t\right)}{dt}\right|$

$\overrightarrow{E}=\frac{{\mu }_{0}n{I}_{max}\omega }{2}r\sin \omega t$