Question

The magnetic flux linked with a circuit changes ${\phi }_i=-5mWbto{\phi }_f=3mWb$in a time interval of 0.03 s. Determine the average emf induced in the circuit for this interval. If the resistance of the circuit is 80 ohm. Determine the average induced current in the circuit.

Answer 1

Step 1: Given data and assumptions

${\phi }_i=-5mWbto{\phi }_f=3mWb$

Time interval$\left(t\right)=0.03s$

Resistance$\left(R\right)=80\Omega$

The average emf induced in the circuit is given as:

$e_{av}=-\left[\frac{{\phi }_f-{\phi }_i}{∆t}\right]$

Where,

$$\begin{gathered} {\phi }_f=\mathrm{Final}\mathrm{linked}\mathrm{flux} \\ {\phi }_f=\mathrm{initial}\mathrm{linked}\mathrm{flux} \end{gathered}$$

$∆t=\mathrm{Change}\mathrm{in}\mathrm{time}\mathrm{interval}$

Step 2: Calculation of the average emf

By substituting the above values in the formula, we get:

$$\begin{gathered} e_{av}=-\left[\frac{3-(-5)}{0.03}\right] \\ {e}_{av}=-\left[\frac{3-(-5)}{0.03}\times \frac{{10}^2}{{10}^2}\right] \\ {e}_{av}=-\left[\frac{800}{3}\right]\mathrm{mV} \end{gathered}$$

The average emf induced in the circuit for the given interval is: $e_{av}=-\left[\frac{800}{3}\right]$

Step 3: Calculation of the current

The formula for average induced current in the circuit

$i=\frac{e_{av}}{R}$

By substituting the above-given values in the above formula, we get:

$$\begin{gathered} i=\frac{800}{3\times 80} \\ i=\frac{10}{3}\mathrm{mA} \end{gathered}$$

Step 4: Final answer

Hence, the average induced current in the circuit is: $i=\frac{10}{3}\mathrm{mA}$