91Ó°ÊÓ

It is given that a straight solenoid having a cross section area of $A=8.00c{m}^2$, it wound with turns of wire per cm hence $n=90/0.01m$which equals to $9000{m^{-}}^1$. And carries a current of and the magnetic field of $\mathrm{Δ}t=0.0400s$the solenoid is turned off hence it becomes zero at a time. It is required to calculate the emf in the second winding of, that encircles the solenoid at the centre.

Now, the induced emf in the winding is given by Faradays law:

$\left|{\mathrm{Î\mu }}_{a_v}\right|=N\left|\frac{\mathrm{Δ}{\mathrm{Ï•}}_B}{\mathrm{Δ}T}\right|$

Now, the initial magnetic flux is$9.50×{{10}^{-}}^{4}V$ and the final magnetic flux is ${\mathrm{ϕ}}_B=B_{f}A$.

The magnetic field of the large straight solenoid is:

$B={\mathrm{μ}}_{0}nI$

Therefore, the average induced emf is:

$$\begin{gathered} \mathrm{Î\mu }=\left|\frac{NA(B_f-B_i)}{\mathrm{Δ}t}\right| \\ =\frac{NA{\mathrm{μ}}_{0}nI}{\mathrm{Δ}t} \\ {\mathrm{Î\mu }}_{a_v}=\frac{\mathrm{μ}012(8.00×{10}^{-}4m^2)\left(9000m^{-}1\right)(0.350A)}{0.0400s} \\ =9.50×{10}^{-}4V \end{gathered}$$

Therefore, the induced emf for the solenoid is$9.50×{{10}^{-}}^{4}V$