A closely wound solenoid of 2000 turns and area of cross-section `1.5xx10^(-4)m^(2)` carries a current of 2.0 a. it suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field `5xx10^(-2)` tesla making an angle of `30^(@)` with the axis of the solenoid. The torque on the solenoid will be:

Options

A. `1.5xx10^(-3)N.m`
B. `1.5xx10^(-2)N.m`
C. `3xx10^(-2)N.m`
D. `3xx10^(-3)N.m`

Answer

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

### Step 1: Calculate the Magnetic Moment (μ)
The magnetic moment (μ) of the solenoid can be calculated using the formula:
\[ \mu = n \cdot I \cdot A \]
Where:
- \( n = 2000 \) (number of turns)
- \( I = 2.0 \, \text{A} \) (current)
- \( A = 1.5 \times 10^{-4} \, \text{m}^2 \) (cross-sectional area)

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

### Step 2: Calculate the Torque (τ)
The torque (τ) on the solenoid in a magnetic field can be calculated using the formula:
\[ \tau = \mu \cdot B \cdot \sin(\theta) \]
Where:
- \( B = 5 \times 10^{-2} \, \text{T} \) (magnetic field strength)
- \( \theta = 30^\circ \)

First, we need to find \( \sin(30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

Now substituting the values into the torque formula:
\[ \tau = 0.6 \cdot (5 \times 10^{-2}) \cdot \left(\frac{1}{2}\right) \]
\[ \tau = 0.6 \cdot 5 \times 10^{-2} \cdot 0.5 \]
\[ \tau = 0.6 \cdot 2.5 \times 10^{-2} \]
\[ \tau = 1.5 \times 10^{-2} \, \text{N m} \]

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

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