Chapter 14, Problem 14.1CYU

Check Your Understanding. A current $I\left(t\right)=\left(5.0A\right)\sin \left(\left(200\pi rad/s\right)t\right)$ flows through the solenoid of part (b) of Example 14.1. What is the maximum emf induced in the surrounding coil?

To determine

The maximum emf induced in the surrounding coil.

Answer to Problem 14.1CYU

The maximum emf induced in the surrounding coil is $\text{4}\text{.56}{\text{×10}}^{\text{-2}}\text{volt}$.

Explanation of Solution

Given info:

  $N_1=500\text{turns}$

  $N_2=10\text{turns}$

  $R_1=3.10\text{cm = 0}\text{.0310 m}$

  $l_1=75\text{cm = 0}\text{.750 m}$

  $I\left(t\right)=\left(5.0\text{A}\right)\sin \left(\left(120\pi rad/s\right)t\right)$

Formula used:

The formula for calculating induced emf in a surrounding coil is:

  $\epsilon =-M\frac{d{I}_1}{dt}$ volt....... (1)

Here, $M$ is mutual inductance and $M=\frac{{\mu }_{\text{o}}{N}_1{N}_2\pi R_1{}^2}{l_1}$....... (2)

Calculation:

From equation (2) we get,

  $\begin{array}{c}M=\frac{{\mu }_{\text{o}}{N}_1{N}_2\pi R_1{}^2}{l_1}\\ =\frac{\left(4\pi \times {10}^{-7}H/m\right)\times 500\times 10\times 3.14\times {\left(0.0310m\right)}^2}{0.750m}\\ =2.53\times {10}^{-5}H\end{array}$

Now, the emf induced in the surrounding coil is,

  $\begin{array}{c}\epsilon =-M\frac{d{I}_1}{dt}\left(\text{taking}\text{the}\text{magnitude}\text{only}\right)\\ =\left(2.53\times {10}^{-5}H\right)\times \frac{d}{dt}\left[\left(5.0A\right)\sin ((120\pi \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)t)\right]\\ =\left(2.53\times {10}^{-5}\right)\times \left(5.0A\right)\times \left(120\pi \right)\times \cos (120\pi \raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.)\\ =2.53\times {10}^{-5}\times 1802.67\\ =0.0456\\ =\text{4}\text{.56}{\text{×10}}^{\text{-2}}\text{volt}\end{array}$

Conclusion:

The maximum emf induced in the surrounding coil is $\text{4}\text{.56}{\text{×10}}^{\text{-2}}\text{volt}$.