A circular coil of radius 4 cm has 50 turns. In this coil a current of 2 A is flowing. It is placed in a magnetic field of `0.1 weber//m^(2).` The amount of work done in rotating it through `180^(@)` from its equilibrium position will be

Options

A. `0.1J`
B. `0.2J`
C. `0.4J`
D. `0.8J`

Answer

To solve the problem step by step, we need to calculate the work done in rotating the circular coil through 180 degrees from its equilibrium position.

### Step 1: Understand the given parameters
- Radius of the coil, \( r = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Number of turns, \( n = 50 \)
- Current flowing through the coil, \( I = 2 \, \text{A} \)
- Magnetic field strength, \( B = 0.1 \, \text{Wb/m}^2 \)

### Step 2: 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 \( r \):
\[ A = \pi (0.04)^2 = \pi (0.0016) \approx 0.005026 \, \text{m}^2 \]

### Step 3: Calculate the magnetic moment \( M \)
The magnetic moment \( M \) of the coil is given by the formula:
\[ M = n \cdot I \cdot A \]
Substituting the values we have:
\[ M = 50 \cdot 2 \cdot 0.005026 \approx 0.5026 \, \text{A m}^2 \]

### Step 4: Calculate the work done \( W \)
The work done in rotating the coil from angle \( \theta_1 \) to \( \theta_2 \) is given by:
\[ W = M \cdot B \cdot (\cos \theta_1 - \cos \theta_2) \]
Here, the coil is rotated from \( \theta_1 = 0^\circ \) to \( \theta_2 = 180^\circ \). Thus:
\[ \cos(0^\circ) = 1 \quad \text{and} \quad \cos(180^\circ) = -1 \]
Substituting these values into the work done formula:
\[ W = M \cdot B \cdot (1 - (-1)) = M \cdot B \cdot 2 \]
Now substituting the values of \( M \) and \( B \):
\[ W = 0.5026 \cdot 0.1 \cdot 2 = 0.10052 \, \text{J} \]

### Step 5: Round off the answer
Rounding off to two decimal places, the work done is approximately:
\[ W \approx 0.1 \, \text{J} \]

### Final Answer
The amount of work done in rotating the coil through \( 180^\circ \) from its equilibrium position is \( 0.1 \, \text{J} \).

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