A coil of `40 Omega` resistance has `100` turns and radius `6 mm` us 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. 2.55 T
D. 0.566 T

Explanation

To find the intensity of the magnetic field when a coil is taken out of the field, we can use the relationship between induced EMF, resistance, charge, and the area of the coil. Here’s a step-by-step solution:

### Step 1: Calculate the Total Resistance
The total resistance \( R_{total} \) in the circuit is the sum of the coil's resistance and the ammeter's resistance.

\[ R_{total} = R_{coil} + R_{ammeter} = 40 \, \Omega + 160 \, \Omega = 200 \, \Omega \]

**Hint:** Remember to add the resistances in series.

### Step 2: Use Faraday's Law of Electromagnetic Induction
According to Faraday's law, the induced EMF (\( \mathcal{E} \)) in the coil is given by:

\[ \mathcal{E} = -\frac{d\Phi}{dt} \cdot N \]

Where:
- \( \Phi \) is the magnetic flux,
- \( N \) is the number of turns in the coil.

Since the coil is taken out of the magnetic field, the initial magnetic flux is \( 0 \) and the final flux is determined by the magnetic field \( B \) and the area \( A \) of the coil.

### Step 3: Calculate the Area of the Coil
The area \( A \) of the coil can be calculated using the formula for the area of a circle:

\[ A = \pi r^2 \]

Given the radius \( r = 6 \, \text{mm} = 6 \times 10^{-3} \, \text{m} \):

\[ A = \pi (6 \times 10^{-3})^2 = \pi (36 \times 10^{-6}) = 36\pi \times 10^{-6} \, \text{m}^2 \]

**Hint:** Convert all measurements to SI units before performing calculations.

### Step 4: Relate Induced EMF to Charge and Resistance
The induced EMF can also be expressed in terms of the charge \( Q \) that flows through the circuit and the total resistance:

\[ \mathcal{E} = I \cdot R_{total} = \frac{Q}{\Delta t} \cdot R_{total} \]

Where \( Q = 32 \, \mu C = 32 \times 10^{-6} \, C \).

### Step 5: Calculate the Current
Since we are not given the time \( \Delta t \), we can rearrange the equation to find the induced EMF:

\[ \mathcal{E} = \frac{Q}{\Delta t} \cdot R_{total} \]

### Step 6: Substitute into Faraday's Law
From Faraday's law, we have:

\[ \mathcal{E} = \frac{N \cdot \Delta \Phi}{\Delta t} \]

Where \( \Delta \Phi = B \cdot A \). Therefore, we can equate:

\[ \frac{Q}{\Delta t} \cdot R_{total} = N \cdot B \cdot A \]

### Step 7: Solve for the Magnetic Field \( B \)
Rearranging gives:

\[ B = \frac{Q \cdot R_{total}}{N \cdot A} \]

Substituting the values:

- \( Q = 32 \times 10^{-6} \, C \)
- \( R_{total} = 200 \, \Omega \)
- \( N = 100 \)
- \( A = 36\pi \times 10^{-6} \, m^2 \)

\[ B = \frac{32 \times 10^{-6} \cdot 200}{100 \cdot 36\pi \times 10^{-6}} \]

### Step 8: Simplify the Expression
\[ B = \frac{6400 \times 10^{-6}}{3600\pi \times 10^{-6}} = \frac{6400}{3600\pi} = \frac{8}{9\pi} \, T \]

### Final Answer
Thus, the intensity of the magnetic field \( B \) is:

\[ B \approx \frac{8}{9\pi} \, T \]

**Hint:** Ensure to simplify fractions carefully and check units for consistency.