Question

A long, straight solenoid has $\mathbf{800}$turns. When the current in the solenoid is $\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{}\mathit{A}$, the average flux through each turn of the solenoid is $\mathbf{3}\mathbf{.}\mathbf{25}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathit{W}\mathit{b}$. What must be the magnitude of the rate of change of the current in order for the self-induced emf to equal $\mathbf{6}\mathbf{.}\mathbf{2}\mathit{m}\mathbf{}\mathit{V}$?

Short answer

The rate of change of current is $8.37\times {10}^{-3}A/s$.

Step-by-step solution

The formula to calculate the EMF induced

The formula to calculate the induced EMF is $\mathbf{\left|}\mathit{\epsilon }\mathbf{\right|}\mathbf{=}\left|-L\frac{di}{dt}\right|$.

All the values provided in the problem are positive. So, do not need to worry about the negative expression.

Calculate the rate of change of current

Given that $\epsilon =6.2mV,I=2.9A,N=800\mathrm{and}{\varphi }_{B_{Aug}}=3.25\times {10}^{-3}Wb.$.

$$\begin{gathered} \frac{di}{dt}=\frac{\epsilon }{L} \\ \frac{di}{dt}=\frac{\epsilon }{{\displaystyle \frac{N{\varphi }_{B_{Aug}}}{I}}} \\ \frac{di}{dt}=\frac{6.2\mathrm{x}{10}^{-3}\mathrm{x}2.9}{800\mathrm{x}3.25\mathrm{x}{10}^{-3}} \\ \frac{di}{dt}=8.37\mathrm{x}{10}^{-3}A/s \end{gathered}$$

So, the rate of change of current is $8.37\mathrm{x}{10}^{-3}A/s$.