A closely wound solenoid of 1000 turns and area of cross-section `2xx10^(-4)m^(2)` carries a current of 1A.It is placed in horizontal axis at`30^(@)` with the diretion of uniform magnetic field of 0.16 T.Magnetic moment of solenoid and torque experienced by the solenoid due to the field are

Options

A. `0.1 Am^(2)`,0.016 Nm
B. `0.2Am^(2)`,0.026 Nm
C. `0.2 Am^(2)`,0.016Nm
D. `0.4 Am^(2)`,0.116Nm

Explanation

To solve the problem, we need to calculate the magnetic moment of the solenoid and the torque experienced by it in the magnetic field. Let's break it down step by step.

### Step 1: Calculate the Magnetic Moment (M)

The formula for the magnetic moment \( M \) of a solenoid is given by:

\[ M = N \cdot I \cdot A \]

Where:
- \( N \) = Number of turns of the solenoid = 1000 turns
- \( I \) = Current through the solenoid = 1 A
- \( A \) = Area of cross-section of the solenoid = \( 2 \times 10^{-4} \, m^2 \)

Now, substituting the values into the formula:

\[ M = 1000 \cdot 1 \cdot (2 \times 10^{-4}) = 1000 \cdot 2 \times 10^{-4} = 0.2 \, A \cdot m^2 \]

### Step 2: Calculate the Torque (\( \tau \))

The torque \( \tau \) experienced by a magnetic dipole in a magnetic field is given by the formula:

\[ \tau = M \cdot B \cdot \sin(\theta) \]

Where:
- \( B \) = Magnetic field strength = 0.16 T
- \( \theta \) = Angle between the magnetic moment and the magnetic field = \( 30^\circ \)

Now, substituting the values into the torque formula:

\[ \tau = 0.2 \cdot 0.16 \cdot \sin(30^\circ) \]

Since \( \sin(30^\circ) = 0.5 \):

\[ \tau = 0.2 \cdot 0.16 \cdot 0.5 = 0.2 \cdot 0.08 = 0.016 \, N \cdot m \]

### Final Answers:

- Magnetic Moment \( M = 0.2 \, A \cdot m^2 \)
- Torque \( \tau = 0.016 \, N \cdot m \)