A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2A. A uniform horizontal magnetic field of strength 2T is turned on. The axis of the coil is initially in the direction of the field . The coil rotates through an angle of `90^(@)` under the influence of the magnetic field. The moment of inertia of the coil is 0.1 kg `m^(2)`
What is the angular speed acquired by the coil when it has rotated by `90^(@)` ?

Options

A. `20 s^(-1)`
B. `30 s^(-1)`
C. `40 s^(-1)`
D. `60 s^(-1)`

Answer

To solve the problem step by step, we will use the concepts of torque, angular acceleration, and the relationship between magnetic moment and angular speed.

### Step 1: Identify the Given Data
- Number of turns, \( N = 100 \)
- Radius of the coil, \( R = 10 \, \text{cm} = 0.1 \, \text{m} \)
- Current, \( I = 3.2 \, \text{A} \)
- Magnetic field strength, \( B = 2 \, \text{T} \)
- Moment of inertia, \( I_{coil} = 0.1 \, \text{kg m}^2 \)
- Angle of rotation, \( \theta = 90^\circ = \frac{\pi}{2} \, \text{radians} \)

### Step 2: Calculate the Area of the Coil
The area \( A \) of the circular coil can be calculated using the formula:
\[ A = \pi R^2 \]
Substituting the value of \( R \):
\[ A = \pi (0.1)^2 = \pi \times 0.01 = 0.01\pi \, \text{m}^2 \]

### Step 3: Calculate the Magnetic Moment
The magnetic moment \( M \) of the coil is given by:
\[ M = N \cdot I \cdot A \]
Substituting the values:
\[ M = 100 \cdot 3.2 \cdot (0.01\pi) = 3.2\pi \, \text{A m}^2 \]

### Step 4: Calculate the Torque Acting on the Coil
The torque \( \tau \) acting on the coil when it is in a magnetic field is given by:
\[ \tau = M \cdot B \cdot \sin(\theta) \]
At \( \theta = 90^\circ \), \( \sin(90^\circ) = 1 \):
\[ \tau = M \cdot B \]
Substituting the values:
\[ \tau = (3.2\pi) \cdot 2 = 6.4\pi \, \text{N m} \]

### Step 5: Relate Torque to Angular Acceleration
Using the relation between torque, moment of inertia, and angular acceleration:
\[ \tau = I_{coil} \cdot \alpha \]
Where \( \alpha \) is the angular acceleration. Rearranging gives:
\[ \alpha = \frac{\tau}{I_{coil}} = \frac{6.4\pi}{0.1} = 64\pi \, \text{rad/s}^2 \]

### Step 6: Use Angular Kinematics to Find Angular Speed
Using the angular kinematic equation:
\[ \omega^2 = \omega_0^2 + 2\alpha \theta \]
Where \( \omega_0 = 0 \) (initial angular speed) and \( \theta = \frac{\pi}{2} \):
\[ \omega^2 = 0 + 2(64\pi)\left(\frac{\pi}{2}\right) \]
\[ \omega^2 = 64\pi^2 \]
Taking the square root:
\[ \omega = 8\pi \, \text{rad/s} \]

### Step 7: Calculate the Numerical Value of Angular Speed
Using \( \pi \approx 3.14 \):
\[ \omega \approx 8 \cdot 3.14 \approx 25.12 \, \text{rad/s} \]

### Final Answer
The angular speed acquired by the coil when it has rotated by \( 90^\circ \) is approximately:
\[ \omega \approx 25.12 \, \text{rad/s} \]