A coil having 30 tums of wire, each of area' `10 cm^(2)`, is placed with its plane at rigiht angle to a magnetic ficld of `0.1 T`. When the coil is suddenly withdrawn from the field, a gaivanometer in series with the coil indicates that a charge of `10^(-5)C` passes around the circuit. What is the combined resistânce (in ohm) of the coil and the galvanometer?

Answer

To solve the problem, we need to find the combined resistance of the coil and the galvanometer using the information provided.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of turns in the coil, \( N = 30 \)
- Area of each turn, \( A = 10 \, \text{cm}^2 = 10 \times 10^{-4} \, \text{m}^2 = 10^{-3} \, \text{m}^2 \)
- Magnetic field strength, \( B = 0.1 \, \text{T} \)
- Charge that passes through the circuit, \( q = 10^{-5} \, \text{C} \)

2. **Calculate the Change in Magnetic Flux (\( \Delta \Phi \)):**
The change in magnetic flux (\( \Delta \Phi \)) when the coil is withdrawn from the magnetic field can be calculated using the formula:
\[ \Delta \Phi = N \cdot B \cdot A \]
Substituting the values:
\[ \Delta \Phi = 30 \cdot 0.1 \cdot 10^{-3} \]
\[ \Delta \Phi = 30 \cdot 0.1 \cdot 10^{-3} = 3 \times 10^{-3} \, \text{Wb} \]

3. **Use the Formula Relating Charge, Change in Flux, and Resistance:**
The relationship between charge (\( q \)), change in magnetic flux (\( \Delta \Phi \)), and resistance (\( R \)) is given by:
\[ q = \frac{\Delta \Phi}{R} \]
Rearranging for \( R \):
\[ R = \frac{\Delta \Phi}{q} \]

4. **Substituting the Values to Find Resistance:**
Substituting the values of \( \Delta \Phi \) and \( q \):
\[ R = \frac{3 \times 10^{-3}}{10^{-5}} \]
\[ R = 3 \times 10^{2} = 300 \, \Omega \]

### Final Answer:
The combined resistance of the coil and the galvanometer is \( 300 \, \Omega \).