A rectangul ar coil of wire of 100 turns and `10xx15cm^(2)` size carrying a current of 2Amp. is in a magnetic field of induction `2 xx 10^(-3)wb//m^(2)`. If the normal drawn to the plane of the coil makes an angle 30° with the field, then the torque on the coil is

Options

A. `3xx10^(-5)N-m`
B. `3xx10^(_3)N-m`
C. `3sqrt3xx10^(-5)N-m`
D. `3sqrt3xx10^(-3)N-m`

Answer

To find the torque on a rectangular coil in a magnetic field, we can use the formula:

\[ \tau = n \cdot i \cdot A \cdot B \cdot \sin(\theta) \]

where:
- \(\tau\) is the torque,
- \(n\) is the number of turns in the coil,
- \(i\) is the current flowing through the coil,
- \(A\) is the area of the coil,
- \(B\) is the magnetic field induction,
- \(\theta\) is the angle between the normal to the coil and the magnetic field.

### Step 1: Identify the given values
- Number of turns, \(n = 100\)
- Current, \(i = 2 \, \text{A}\)
- Area of the coil, \(A = 10 \, \text{cm} \times 15 \, \text{cm} = 150 \, \text{cm}^2 = 150 \times 10^{-4} \, \text{m}^2\)
- Magnetic field induction, \(B = 2 \times 10^{-3} \, \text{Wb/m}^2\)
- Angle, \(\theta = 30^\circ\)

### Step 2: Convert area to square meters
\[ A = 150 \, \text{cm}^2 = 150 \times 10^{-4} \, \text{m}^2 = 1.5 \times 10^{-2} \, \text{m}^2 \]

### Step 3: Calculate \(\sin(\theta)\)
For \(\theta = 30^\circ\):
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 4: Substitute values into the torque formula
Now, substituting the values into the torque formula:
\[ \tau = n \cdot i \cdot A \cdot B \cdot \sin(\theta) \]
\[ \tau = 100 \cdot 2 \cdot (1.5 \times 10^{-2}) \cdot (2 \times 10^{-3}) \cdot \left(\frac{1}{2}\right) \]

### Step 5: Simplify the expression
Calculating step by step:
1. Calculate \(n \cdot i = 100 \cdot 2 = 200\)
2. Calculate \(A \cdot B = (1.5 \times 10^{-2}) \cdot (2 \times 10^{-3}) = 3 \times 10^{-5}\)
3. Now, multiply:
\[ \tau = 200 \cdot 3 \times 10^{-5} \cdot \frac{1}{2} \]
\[ \tau = 200 \cdot 1.5 \times 10^{-5} = 3 \times 10^{-3} \, \text{N m} \]

### Final Answer
The torque on the coil is:
\[ \tau = 3 \times 10^{-3} \, \text{N m} \]

---