Calculating Induced EMF Using Faraday's Law

This problem involves calculating the magnitude of the electromotive force (EMF) induced in a coil due to a change in magnetic flux. The fundamental principle governing this phenomenon is Faraday's Law of Electromagnetic Induction.

Given Information

• Number of turns in the coil, N = 500
• Initial magnetic flux, $\Phi_{B,i} = 0.2 \text{ mWb}$
• Final magnetic flux, $\Phi_{B,f} = 0.8 \text{ mWb}$
• Time interval for the change, $\Delta t = 0.02 \text{ s}$

Faraday's Law Formula

Faraday's Law states that the induced EMF ($\mathcal{E}$) in a coil is proportional to the number of turns (N) and the rate of change of magnetic flux ($\frac{\Delta\Phi_B}{\Delta t}$). The magnitude of the induced EMF is given by:

Step-by-Step Calculation

1. Calculate the change in magnetic flux ($\Delta\Phi_B$):

   The change in flux is the difference between the final and initial flux values.

   $ \Delta\Phi_B = \Phi_{B,f} - \Phi_{B,i} $ $ \Delta\Phi_B = 0.8 \text{ mWb} - 0.2 \text{ mWb} = 0.6 \text{ mWb} $
2. Convert magnetic flux to Webers (Wb):

   Since $1 \text{ mWb} = 10^{-3} \text{ Wb}$, we convert the change in flux:

   $ \Delta\Phi_B = 0.6 \times 10^{-3} \text{ Wb} $
3. Calculate the rate of change of magnetic flux:

   Divide the change in flux by the time interval.

   $ \frac{\Delta\Phi_B}{\Delta t} = \frac{0.6 \times 10^{-3} \text{ Wb}}{0.02 \text{ s}} $ $ \frac{\Delta\Phi_B}{\Delta t} = \frac{0.6}{0.02} \times 10^{-3} \text{ Wb/s} $ $ \frac{\Delta\Phi_B}{\Delta t} = 30 \times 10^{-3} \text{ V} $

   (Note: $1 \text{ Wb/s} = 1 \text{ Volt (V)}$)

4. Calculate the magnitude of the induced EMF ($\mathcal{E}$):

   Multiply the rate of change of flux by the number of turns.

   $ |\mathcal{E}| = N \times \left| \frac{\Delta\Phi_B}{\Delta t} \right| $ $ |\mathcal{E}| = 500 \times (30 \times 10^{-3} \text{ V}) $ $ |\mathcal{E}| = 500 \times 0.030 \text{ V} $ $ |\mathcal{E}| = 15 \text{ V} $

Conclusion

The magnitude of the EMF induced in the coil is 15 V.