A coil having 1,000 turns closely wound and an area of cross section `2 xx 10^(-5) m^(2)` carries a current of 4A. The coil is kept in a region of uniform magnetic field of` 0.2` T. Calculate the net torque acting on the coil if it is free to move about vertical axis and its axis makes an angle of `30^(@)` with the magnetic field lines.

Answer

To calculate the net torque acting on the coil, we can use the formula for torque (\( \tau \)) acting on a current-carrying coil in a magnetic field:

\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \]

Where:
- \( N \) = number of turns of the coil
- \( I \) = current flowing through the coil (in Amperes)
- \( A \) = area of cross-section of the coil (in square meters)
- \( B \) = magnetic field strength (in Teslas)
- \( \theta \) = angle between the normal to the coil and the magnetic field (in degrees)

### Step-by-step Solution:

1. **Identify the given values:**
- Number of turns, \( N = 1000 \)
- Current, \( I = 4 \, A \)
- Area, \( A = 2 \times 10^{-5} \, m^2 \)
- Magnetic field strength, \( B = 0.2 \, T \)
- Angle, \( \theta = 30^\circ \)

2. **Convert the angle to radians if necessary:**
- For this calculation, we can use the sine of the angle directly in degrees.

3. **Calculate \( \sin(30^\circ) \):**
- \( \sin(30^\circ) = \frac{1}{2} \)

4. **Substitute the values into the torque formula:**
\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \]
\[ \tau = 1000 \cdot 4 \cdot (2 \times 10^{-5}) \cdot 0.2 \cdot \sin(30^\circ) \]
\[ \tau = 1000 \cdot 4 \cdot (2 \times 10^{-5}) \cdot 0.2 \cdot \frac{1}{2} \]

5. **Simplify the expression:**
\[ \tau = 1000 \cdot 4 \cdot (2 \times 10^{-5}) \cdot 0.2 \cdot 0.5 \]
\[ \tau = 1000 \cdot 4 \cdot (2 \times 10^{-5}) \cdot 0.1 \]
\[ \tau = 1000 \cdot 4 \cdot 2 \times 10^{-6} \]
\[ \tau = 8000 \times 10^{-6} \, N \cdot m \]
\[ \tau = 0.008 \, N \cdot m \]

6. **Final Result:**
The net torque acting on the coil is \( \tau = 0.008 \, N \cdot m \).