Question

A solenoid with $600$ turns per metre and a radius of $2$ $cm$, carries a time varying current $i\left(t\right)=(6+4t^2$) A. The electric field at a distance $4$ $cm$ from the axis of the solenoid at $t=2$ $s$ will be $($in $\mu$V m${}^{-1}$ to the nearest integer )

• A. $48$
• B. $60$
• C. $68$
• D. $38$

Answer 1

Answer: (B)

$60$

Given
$i=(6+4t^2)amp$
Magnetic field for a solenoid: $B={\mu }_{0}ni={\mu }_{0}n(6+4t^2)$
Flux passing through a circle of radius 4 : $\varphi =\pi r^{2}B$
Magnetic field outside the solenoid of radius 2 cm is zero.
$\therefore \varphi =\pi r^{2}B={\mu }_0\times \left(600\right)\pi \times 2^2\times {10}^{-4}(6+4t^2)$

$\frac{d\varphi }{dt}=(4\pi \times {10}^{-7})\times \left(600\right)\times (\pi \times 4\times {10}^{-4})\times 8t=$Induced emf in a circle of radius 4 cm:

$\epsilon =\oint \overrightarrow{E}.\overrightarrow{dl}=E\times 2\pi \times 4\times {10}^{-2}=\frac{d\varphi }{dt}=4\times {10}^{-4}\pi \times 8t$

$\therefore E=\frac{\frac{d\varphi }{dt}}{2\pi \times 4\times {10}^{-2}}=\frac{(4\pi \times {10}^{-7})\times \left(600\right)\times (\pi \times 4\times {10}^{-4})\times 8t}{2\pi \times 4\times {10}^{-2}}$

$\Rightarrow E=16\pi \times 6\times {10}^{-7}\times t$
Putting $t=2$ we get

$\Rightarrow E=16\pi \times 6\times 2\times {10}^{-7}=603\times {10}^{-7}V/m=6.03\times {10}^{-5}V/m$

$\Rightarrow E=60.3\times {10}^{-6}V/m=60.3\mu V/m$