Question

A long, straight solenoid with a cross-sectional area of 8.00 cm2 is wound with 90 turns of wire per centimetre, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its centre. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

Short answer

Answer:

The average induced emf in the second winding if the current in the solenoid is turned off in such a way that the magnetic field of the solenoid becomes zero in 0.400 s is $9.50\times {{10}^{-}}^{4}V$

Step-by-step solution

Average induced emf

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 $\Delta t=0.0400s$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:

$\left|{\epsilon }_{a_v}\right|=N\left|\frac{\Delta {\varphi }_B}{\Delta T}\right|$

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

The magnetic field of the large straight solenoid is:

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

Calculating the value of the induced emf

Therefore, the average induced emf is:

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

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