A circular coil opf `500` turns of wire has an enclosed area of `0.1 m^(2)` per turn. It is kept perpendicular to a magnetic field of induction `0.2 T` and rotated by `180^(@)` about a diameter perpendicular to the field in `0.1` sec. how much charge will pass when the coil is connected to a gavanometer with a combined resistance of `50 ohms`

Options

A. `0.2 C`
B. `0.4 C`
C. `2 C`
D. `4 C`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the initial magnetic flux (Φ_initial)
The magnetic flux (Φ) through a coil is given by the formula:
\[ \Phi = B \cdot A \cdot \cos(\theta) \]
where:
- \( B \) is the magnetic field strength (0.2 T),
- \( A \) is the area of the coil (0.1 m²),
- \( \theta \) is the angle between the magnetic field and the normal to the coil's surface.

Initially, the coil is perpendicular to the magnetic field, so \( \theta = 0^\circ \):
\[ \Phi_{\text{initial}} = B \cdot A \cdot \cos(0) = 0.2 \cdot 0.1 \cdot 1 = 0.02 \, \text{Wb} \]

### Step 2: Calculate the final magnetic flux (Φ_final)
After rotating the coil by \( 180^\circ \), the angle becomes \( \theta = 180^\circ \):
\[ \Phi_{\text{final}} = B \cdot A \cdot \cos(180) = 0.2 \cdot 0.1 \cdot (-1) = -0.02 \, \text{Wb} \]

### Step 3: Calculate the change in magnetic flux (ΔΦ)
The change in magnetic flux (ΔΦ) is given by:
\[ \Delta \Phi = \Phi_{\text{final}} - \Phi_{\text{initial}} \]
Substituting the values:
\[ \Delta \Phi = -0.02 - 0.02 = -0.04 \, \text{Wb} \]

### Step 4: Calculate the induced EMF (ε)
The induced EMF (ε) in the coil can be calculated using Faraday's law of electromagnetic induction:
\[ \varepsilon = -N \frac{\Delta \Phi}{\Delta t} \]
where:
- \( N \) is the number of turns (500),
- \( \Delta t \) is the time taken for the change (0.1 s).

Substituting the values:
\[ \varepsilon = -500 \cdot \frac{-0.04}{0.1} = 500 \cdot 0.4 = 200 \, \text{V} \]

### Step 5: Calculate the charge (Q) passing through the galvanometer
Using Ohm's law, the relationship between charge, current, and time is given by:
\[ Q = I \cdot t \]
where \( I \) can be expressed as:
\[ I = \frac{\varepsilon}{R} \]
Substituting the values:
\[ I = \frac{200}{50} = 4 \, \text{A} \]
Now, substituting \( I \) back to find \( Q \):
\[ Q = I \cdot t = 4 \cdot 0.1 = 0.4 \, \text{C} \]

### Final Answer
The charge that will pass through the galvanometer is **0.4 C**.