One coil of resistance `40 Omega` is connected to a galvanometer of `16 Omega` resistance . The coil has radius 6mm and turns 100. This coil is placed between the poles of a magnet such that magnetic field is perpendicular to coil. If coil is dragged out then the charge through the galvanometer is `32 muC`. The magnetic field is :-

Options

A. 6.55 T
B. 5.66 T
C. 0.655 T
D. 0.566 T

Answer

To find the magnetic field \( B \) when a coil is dragged out of a magnetic field, we can use the relationship between induced electromotive force (emf), charge, resistance, and the number of turns in the coil. Here’s how we can solve the problem step by step:

### Step 1: Identify the given values
- Resistance of the coil, \( R_c = 40 \, \Omega \)
- Resistance of the galvanometer, \( R_g = 16 \, \Omega \)
- Total resistance, \( R = R_c + R_g = 40 \, \Omega + 16 \, \Omega = 56 \, \Omega \)
- Charge through the galvanometer, \( Q = 32 \, \mu C = 32 \times 10^{-6} \, C \)
- Number of turns in the coil, \( N = 100 \)
- Radius of the coil, \( r = 6 \, mm = 6 \times 10^{-3} \, m \)

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

### Step 3: Use the formula for induced emf
The induced emf \( \mathcal{E} \) can be related to the charge \( Q \) and the total resistance \( R \):
\[ \mathcal{E} = \frac{Q}{R} \]
Substituting the values:
\[ \mathcal{E} = \frac{32 \times 10^{-6}}{56} \approx 5.714 \times 10^{-7} \, V \]

### Step 4: Relate induced emf to magnetic field
The induced emf can also be expressed in terms of the magnetic field \( B \), the area \( A \), and the number of turns \( N \):
\[ \mathcal{E} = N \frac{d\Phi}{dt} = N \frac{d(B \cdot A)}{dt} = N A \frac{dB}{dt} \]
Since the coil is dragged out, we can assume the change in magnetic field \( dB \) occurs over a time interval \( dt \), which we can ignore for this calculation as we are looking for \( B \).

### Step 5: Rearranging the equation to solve for \( B \)
Rearranging the equation gives:
\[ B = \frac{\mathcal{E} \cdot R}{N \cdot A} \]
Substituting the values we have:
\[ B = \frac{(5.714 \times 10^{-7}) \cdot 56}{100 \cdot (36\pi \times 10^{-6})} \]

### Step 6: Calculate the magnetic field
Calculating the values:
\[ B \approx \frac{3.2 \times 10^{-5}}{3.6 \times 10^{-4}} \approx 0.566 \, T \]

### Final Answer
The magnetic field \( B \) is approximately \( 0.566 \, T \).

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