Problem

A coil of resistance ( 100 , \Omega ) is placed in a magnetic field of ( 1 , \mathrm{mWb} ). The coil has 100 turns, and a galvanometer of ( 400 , \Omega ) resistance is connected in series with it. Find the average e.m.f. and the current if the coil is moved in ( 1/10 , \mathrm{s} ) from the given field to a field of ( 0.2 , \mathrm{mWb} ).

Data

• Number of turns: ( N = 100 )
• Initial magnetic flux:

$$\Phi_1 = 1 \, \mathrm{mWb} = 1 \times 10^{-3} \, \mathrm{Wb}$$

• Final magnetic flux:

$$\Phi_2 = 0.2 \, \mathrm{mWb} = 0.2 \times 10^{-3} \, \mathrm{Wb}$$

• Resistance of the coil: ( R_1 = 100 , \Omega )
• Resistance of the galvanometer: ( R_2 = 400 , \Omega )
• Time for the change: ( \Delta t = 0.1 , \mathrm{s} )

To find:

1. Average e.m.f. (( \varepsilon )).
2. Current (( I )).

Prerequisite Concepts

1. Faraday’s law of electromagnetic induction:

$$\varepsilon = -N \frac{\Delta \Phi}{\Delta t}.$$

2. Total resistance of the circuit:

$$R_{\text{eq}} = R_1 + R_2.$$

3. Current:

$$I = \frac{\varepsilon}{R_{\text{eq}}}.$$

Solution

Step 1: Calculate the Average e.m.f.

1. Using Faraday’s law:

$$\varepsilon = -N \frac{\Delta \Phi}{\Delta t}.$$

2. Change in magnetic flux:

$$\Delta \Phi = \Phi_2 - \Phi_1 = 0.2 \times 10^{-3} - 1 \times 10^{-3} = -0.8 \times 10^{-3} \, \mathrm{Wb}.$$

3. Substituting values:

$$\varepsilon = -100 \cdot \frac{-0.8 \times 10^{-3}}{0.1}.$$

4. Simplify:

$$\varepsilon = 0.8 \, \mathrm{V}.$$

Step 2: Calculate the Current

1. Total resistance of the circuit:

$$R_{\text{eq}} = R_1 + R_2 = 100 + 400 = 500 \, \Omega.$$

2. Using Ohm’s law:

$$I = \frac{\varepsilon}{R_{\text{eq}}}.$$

3. Substituting values:

$$I = \frac{0.8}{500}.$$

4. Simplify:

$$I = 0.0016 \, \mathrm{A} = 1.6 \, \mathrm{mA}.$$

Answer

1. Average e.m.f.:

$$\varepsilon = 0.8 \, \mathrm{V}.$$

2. Current:

$$I = 1.6 \, \mathrm{mA}.$$

---