A 250-turns rectangular coil of length 2.1 cm and width 1.25 cm carries a current of `85 muA` and subjected to magnetic field of strength `0.85 T`. Work done for rotating the coil by `180^(@)` against the torque is

Options

A. `4.55 muJ`
B. `2.3 muJ`
C. `1.5 mu J`
D. `9.1 mu J`

Explanation

To solve the problem of calculating the work done in rotating a rectangular coil by 180 degrees against the torque, we can follow these steps:

### Step 1: Understand the Formula for Work Done
The work done (W) in rotating a coil in a magnetic field can be expressed as:
\[ W = \mu B (\cos \theta_1 - \cos \theta_2) \]
where:
- \( \mu \) is the magnetic moment of the coil,
- \( B \) is the magnetic field strength,
- \( \theta_1 \) is the initial angle,
- \( \theta_2 \) is the final angle.

### Step 2: Calculate the Magnetic Moment (\( \mu \))
The magnetic moment (\( \mu \)) for a coil is given by:
\[ \mu = n \cdot I \cdot A \]
where:
- \( n \) is the number of turns (250 turns),
- \( I \) is the current (85 µA = \( 85 \times 10^{-6} \) A),
- \( A \) is the area of the coil.

First, we need to calculate the area \( A \):
- Length \( l = 2.1 \) cm = \( 0.021 \) m,
- Width \( w = 1.25 \) cm = \( 0.0125 \) m.

Thus,
\[ A = l \cdot w = 0.021 \, \text{m} \times 0.0125 \, \text{m} = 0.0002625 \, \text{m}^2 \]

Now, substituting the values into the magnetic moment formula:
\[ \mu = 250 \cdot (85 \times 10^{-6}) \cdot 0.0002625 \]

### Step 3: Calculate the Magnetic Moment
Calculating \( \mu \):
\[ \mu = 250 \cdot 85 \times 10^{-6} \cdot 0.0002625 \]
\[ \mu = 250 \cdot 85 \cdot 0.0002625 \times 10^{-6} \]
\[ \mu = 5.515625 \times 10^{-6} \, \text{A m}^2 \]

### Step 4: Substitute Values into the Work Done Formula
Now substituting the values into the work done formula:
- \( B = 0.85 \, \text{T} \)
- \( \theta_1 = 0^\circ \) (initial position)
- \( \theta_2 = 180^\circ \) (final position)

Using \( \cos(0^\circ) = 1 \) and \( \cos(180^\circ) = -1 \):
\[ W = \mu B (\cos 0 - \cos 180) \]
\[ W = \mu B (1 - (-1)) \]
\[ W = \mu B (1 + 1) \]
\[ W = 2 \mu B \]

### Step 5: Calculate the Work Done
Now substituting \( \mu \) and \( B \):
\[ W = 2 \cdot (5.515625 \times 10^{-6}) \cdot 0.85 \]
\[ W = 2 \cdot 5.515625 \times 0.85 \times 10^{-6} \]
\[ W = 9.48 \times 10^{-6} \, \text{J} \]

### Final Answer
The work done for rotating the coil by \( 180^\circ \) against the torque is:
\[ W \approx 9.48 \, \mu J \]

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