A coil 10 turns and a resistance of 20`Omega` is connected in series with B.G. of resistance 30`Omega` . The coil is placed with its plane perpendicular to the direction of a uniform magnetic field of induction `10^(-2)` T. If it is now turned through an angle of `60^@` about an axis in its plane. Find the charge induced in the coil.(Area of a coil= `10^(-2)` `m^2`

Options

A. `2 xx 10^(-5)`C
B. `3.2 xx 10^(-5) C`
C. `1 xx 10^(-5)C`
D. `5.5 xx 10^(-5) C`

Answer

To solve the problem, we need to calculate the charge induced in the coil when it is turned through an angle of \(60^\circ\) in a magnetic field. We will use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a coil is equal to the negative rate of change of magnetic flux through the coil.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Number of turns in the coil, \(n = 10\)
- Resistance of the coil, \(R_1 = 20 \, \Omega\)
- Resistance of the battery (B.G.), \(R_2 = 30 \, \Omega\)
- Total resistance, \(R = R_1 + R_2 = 20 + 30 = 50 \, \Omega\)
- Magnetic field induction, \(B = 10^{-2} \, \text{T}\)
- Area of the coil, \(A = 10^{-2} \, \text{m}^2\)
- Initial angle, \(\theta_1 = 0^\circ\) (since the plane is perpendicular to the magnetic field)
- Final angle, \(\theta_2 = 60^\circ\)

2. **Calculate Initial Magnetic Flux (\(\Phi_1\)):**
\[ \Phi_1 = B \cdot A \cdot \cos(\theta_1) = B \cdot A \cdot \cos(0^\circ) = B \cdot A = (10^{-2}) \cdot (10^{-2}) = 10^{-4} \, \text{Wb} \]

3. **Calculate Final Magnetic Flux (\(\Phi_2\)):**
\[ \Phi_2 = B \cdot A \cdot \cos(\theta_2) = B \cdot A \cdot \cos(60^\circ) = (10^{-2}) \cdot (10^{-2}) \cdot \frac{1}{2} = 5 \times 10^{-5} \, \text{Wb} \]

4. **Calculate Change in Magnetic Flux (\(\Delta \Phi\)):**
\[ \Delta \Phi = \Phi_2 - \Phi_1 = 5 \times 10^{-5} - 10^{-4} = -5 \times 10^{-5} \, \text{Wb} \]

5. **Calculate Induced EMF (\(\mathcal{E}\)):**
Using Faraday's law:
\[ \mathcal{E} = -n \frac{\Delta \Phi}{\Delta t} \quad \text{(assuming \(\Delta t\) is very small, we can treat it as instantaneous)} \]
Since we are not given time, we can directly use the change in flux:
\[ \mathcal{E} = n \frac{-\Delta \Phi}{\Delta t} = n \cdot \frac{5 \times 10^{-5}}{\Delta t} \]

6. **Calculate Induced Charge (\(Q\)):**
The induced charge can be calculated using:
\[ Q = \frac{\mathcal{E} \cdot \Delta t}{R} \]
Substituting for \(\mathcal{E}\):
\[ Q = \frac{n \cdot 5 \times 10^{-5}}{R} \cdot \Delta t \]
Since we are looking for charge induced, we can consider \(\Delta t\) to be 1 second for simplicity:
\[ Q = \frac{10 \cdot 5 \times 10^{-5}}{50} = \frac{5 \times 10^{-4}}{50} = 1 \times 10^{-5} \, \text{C} \]

### Final Answer:
The charge induced in the coil is \(Q = 1 \times 10^{-5} \, \text{C}\).