A coil of `100` turns and area `2 xx 10^(-2) m^(2)`, pivoted about a vertical diameter in a uniform magnetic field carries a current of `5A`. When the coil is held with its plane in North-South direction, it experiences a torque of `0.3 N//m`. When the plane is in East-West direction the torque is `0.4 Nm`. The value of magnetic induction is (Neglect earth's magnetic field)

Options

A. `0.12T`
B. `0.03T`
C. `0.14T`
D. None of these

Answer

To solve the problem, we need to find the value of the magnetic induction (B) given the torque experienced by a coil in two different orientations in a magnetic field.

### Step-by-Step Solution:

1. **Understanding the Torque Formula**:
The torque (\( \tau \)) experienced by a coil in a magnetic field is given by the formula:
\[ \tau = N \cdot I \cdot A \cdot B \cdot \sin(\theta) \]
where:
- \( N \) = number of turns in the coil
- \( I \) = current through the coil
- \( A \) = area of the coil
- \( B \) = magnetic induction
- \( \theta \) = angle between the magnetic field and the normal to the coil's plane

2. **Torque in North-South Direction**:
When the coil is in the North-South direction, the angle \( \theta \) is \( 90^\circ \) (since the area vector is perpendicular to the magnetic field). Therefore, \( \sin(90^\circ) = 1 \):
\[ \tau_1 = N \cdot I \cdot A \cdot B_1 \]
Given \( \tau_1 = 0.3 \, \text{Nm} \), \( N = 100 \), \( I = 5 \, \text{A} \), and \( A = 2 \times 10^{-2} \, \text{m}^2 \):
\[ 0.3 = 100 \cdot 5 \cdot (2 \times 10^{-2}) \cdot B_1 \]
Simplifying this:
\[ 0.3 = 1000 \cdot (2 \times 10^{-2}) \cdot B_1 \]
\[ 0.3 = 20 \cdot B_1 \]
\[ B_1 = \frac{0.3}{20} = 0.015 \, \text{T} \]

3. **Torque in East-West Direction**:
When the coil is in the East-West direction, the angle \( \theta \) is \( 0^\circ \) (the area vector is parallel to the magnetic field). Therefore, \( \sin(0^\circ) = 0 \):
\[ \tau_2 = N \cdot I \cdot A \cdot B_2 \cdot \sin(0^\circ) \]
Given \( \tau_2 = 0.4 \, \text{Nm} \):
\[ 0.4 = N \cdot I \cdot A \cdot B_2 \]
Here, \( \tau_2 = 0.4 \):
\[ 0.4 = 100 \cdot 5 \cdot (2 \times 10^{-2}) \cdot B_2 \]
Simplifying this:
\[ 0.4 = 1000 \cdot (2 \times 10^{-2}) \cdot B_2 \]
\[ 0.4 = 20 \cdot B_2 \]
\[ B_2 = \frac{0.4}{20} = 0.02 \, \text{T} \]

4. **Finding the Resultant Magnetic Field**:
Since the two magnetic fields \( B_1 \) and \( B_2 \) are perpendicular to each other, we can find the resultant magnetic induction \( B \) using the Pythagorean theorem:
\[ B = \sqrt{B_1^2 + B_2^2} \]
Substituting the values:
\[ B = \sqrt{(0.015)^2 + (0.02)^2} \]
\[ B = \sqrt{0.000225 + 0.0004} \]
\[ B = \sqrt{0.000625} = 0.025 \, \text{T} \]

### Final Answer:
The value of magnetic induction \( B \) is \( 0.025 \, \text{T} \).