A coil of `40 Omega` resistance has `100` turns and radius `6 mm` is connected to ammeter of resistance of `160 ohms`. Coil is placed perpendicular to the magnetic field. When coil is taken out of the field, `32 mu C` charge flows through it. The intensity of magnetic field will be

Options

A. `6.55 T`
B. `5.66 T`
C. `0.655 T`
D. `0.565 T`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the given values
- Resistance of the coil, \( R_{coil} = 40 \, \Omega \)
- Number of turns, \( N = 100 \)
- Radius of the coil, \( r = 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m} \)
- Resistance of the ammeter, \( R_{ammeter} = 160 \, \Omega \)
- Total charge that flows, \( Q = 32 \, \mu C = 32 \times 10^{-6} \, C \)

### Step 2: Calculate the total resistance in the circuit
The total resistance \( R_{total} \) in the circuit is the sum of the resistance of the coil and the resistance of the ammeter:
\[ R_{total} = R_{coil} + R_{ammeter} = 40 \, \Omega + 160 \, \Omega = 200 \, \Omega \]

### Step 3: Calculate the area of the coil
The area \( A \) of the circular coil can be calculated using the formula:
\[ A = \pi r^2 \]
Substituting the value of \( r \):
\[ A = \pi (6 \times 10^{-3})^2 = \pi (36 \times 10^{-6}) \approx 1.131 \times 10^{-4} \, m^2 \]

### Step 4: Use the formula for induced charge
The induced charge \( Q \) can be expressed using the formula:
\[ Q = N \cdot \frac{B \cdot A \cdot \Delta \Phi}{R_{total}} \]
Where \( \Delta \Phi \) is the change in magnetic flux. Since the coil is taken out of the magnetic field, we have:
\[ \Delta \Phi = B \cdot A \cdot \cos(0) - 0 = B \cdot A \]
Thus, we can rewrite the equation as:
\[ Q = N \cdot \frac{B \cdot A}{R_{total}} \]

### Step 5: Rearranging to find the magnetic field \( B \)
Rearranging the equation to solve for \( B \):
\[ B = \frac{Q \cdot R_{total}}{N \cdot A} \]

### Step 6: Substitute the values into the equation
Substituting the known values:
\[ B = \frac{(32 \times 10^{-6}) \cdot (200)}{100 \cdot (1.131 \times 10^{-4})} \]
Calculating this:
\[ B = \frac{6.4 \times 10^{-4}}{1.131 \times 10^{-2}} \approx 0.0565 \, T \]

### Final Answer
The intensity of the magnetic field \( B \) is approximately \( 0.0565 \, T \) or \( 56.5 \, mT \).

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