Question

The current in an ideal, long solenoid is varied at a uniform rate of $0.01A/s$. The solenoid has $2000$ turns/m and its radius is $6.0cm$. (a) Consider a circle of radius $1.0cm$ inside the solenoid with its axis coinciding with the axis of the solenoid. Write the change in the magnetic flux through this circle in $2.0$ seconds. (b) Find the electric field induced at a point on the circumference of the circle. (c) Find the electric field induced at a point outside the solenoid at a distance $8.0cm$ from its axis.

Answer 1

$\frac{di}{dt}=0.01A/S$

For $2s\left(\frac{di}{dt}\right)=0.02A/S$

$n=2000$ turn/m, $R=6.0$ cm = $0.06$ m

$r=1$ cm = $0.01$ m

$\begin{array}{c}\left(a\right)\varphi =BA\\ \Rightarrow \frac{d\varphi }{dt}={\mu }_{o}nA\frac{di}{dt}\\ =4\pi \times {10}^{-7}\times 2\times {10}^{-3}\times \pi \times 1\times 10\\ =16{\pi }^2\times {10}^{-10}\omega \\ =1.6\times {10}^{-8}\omega \end{array}$

or, $\frac{d\varphi }{dt}$ for $1s=0.785\omega$

$\begin{array}{c}\left(b\right)\int E.dl=\frac{d\varphi }{dt}\\ \Rightarrow E\varphi dl=\frac{d\varphi }{dt}\Rightarrow E=\frac{0.785\times {10}^{-8}}{2\pi \times {10}^{-2}}=1.2\times {10}^{-7}V/m\\ \left(c\right)\frac{d\varphi }{dt}={\mu }_{o}n\frac{di}{dt}A=4\pi \times {10}^{-7}\times 2000\times 0.01\times \pi \times {\left(0.06\right)}^2\\ E\varphi dl=\frac{d\varphi }{dt}\\ E=\frac{d\varphi /dt}{2\pi r}=\frac{4\pi \times {10}^{-7}\times 2000\times 0.01\times \pi \times {\left(0.06\right)}^2}{\pi \times 8\times {10}^{-2}}=5.64\times {10}^{-7}V/m\end{array}$