A closely wound solenoid of `2000` turns and area of cross-section `1.5 xx 10^(-4) m^(2)` carries a current of `2.0 A`. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field `5 xx 10^(-2)` tesla making an angle of `30^(@)` with the axis of the solenoid. The torque on the solenoid with the

Options

A. `3 xx 10^(-3) N.m`
B. `1.5 xx 10^(-3) N.m`
C. `1.5 xx 10^(-2) N.m`
D. `3 xx 10^(-2) N.m`

Explanation

To find the torque on the solenoid, we can follow these steps:

### Step 1: Calculate the magnetic moment (M) of the solenoid.
The magnetic moment \( M \) of a solenoid is given by the formula:
\[ M = N \cdot I \cdot A \]
where:
- \( N \) = number of turns (2000 turns),
- \( I \) = current (2.0 A),
- \( A \) = area of cross-section of the solenoid (\( 1.5 \times 10^{-4} \, m^2 \)).

Substituting the values:
\[ M = 2000 \cdot 2.0 \cdot 1.5 \times 10^{-4} \]
\[ M = 4000 \cdot 1.5 \times 10^{-4} \]
\[ M = 6.0 \times 10^{-1} \, A \cdot m^2 = 0.6 \, A \cdot m^2 \]

### Step 2: Calculate the torque (τ) on the solenoid.
The torque \( \tau \) on a current-carrying loop in a magnetic field is given by:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
where:
- \( B \) = magnetic field strength (\( 5 \times 10^{-2} \, T \)),
- \( \theta \) = angle between the magnetic moment and the magnetic field (30°).

Substituting the values:
\[ \tau = 0.6 \cdot (5 \times 10^{-2}) \cdot \sin(30^\circ) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \tau = 0.6 \cdot (5 \times 10^{-2}) \cdot \frac{1}{2} \]
\[ \tau = 0.6 \cdot 2.5 \times 10^{-2} \]
\[ \tau = 1.5 \times 10^{-2} \, N \cdot m \]

### Final Answer:
The torque on the solenoid is:
\[ \tau = 1.5 \times 10^{-2} \, N \cdot m \]

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