A current carrying ring with its center at origin and moment of inertia `2xx10^(-2)kg-m^(2)` about an axis passing through its centre and perpendicular to its plane has magnetic moment `M=(3hati-4hatj)` `A-m^(2)`. At time t=0 a magnetic field `B=(4hati+3hatj)` T is switched on.
Angular acceleration of the ring at time t=0, in `rad//s^(2)` is

Options

A. 5000
B. 1250
C. 2500
D. zero

Answer

To solve the problem, we need to find the angular acceleration of a current-carrying ring when a magnetic field is switched on. We will follow these steps:

### Step 1: Identify the Given Values
- Moment of inertia (I) of the ring: \( I = 2 \times 10^{-2} \, \text{kg m}^2 \)
- Magnetic moment (M): \( M = (3\hat{i} - 4\hat{j}) \, \text{A m}^2 \)
- Magnetic field (B): \( B = (4\hat{i} + 3\hat{j}) \, \text{T} \)

### Step 2: Calculate the Torque (τ)
The torque (τ) acting on the ring due to the magnetic field can be calculated using the cross product of the magnetic moment (M) and the magnetic field (B):
\[ \tau = \mathbf{M} \times \mathbf{B} \]

Substituting the values of M and B:
\[ \tau = (3\hat{i} - 4\hat{j}) \times (4\hat{i} + 3\hat{j}) \]

### Step 3: Compute the Cross Product
Using the determinant method for the cross product:
\[ \tau = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 0 \\ 4 & 3 & 0 \end{vmatrix} \]

Calculating the determinant:
\[ \tau = \hat{k}(3 \cdot 3 - (-4) \cdot 4) = \hat{k}(9 + 16) = 25\hat{k} \, \text{N m} \]

### Step 4: Relate Torque to Angular Acceleration
The relationship between torque (τ), moment of inertia (I), and angular acceleration (α) is given by:
\[ \tau = I \alpha \]

### Step 5: Solve for Angular Acceleration (α)
Substituting the values we have:
\[ 25 = (2 \times 10^{-2}) \alpha \]

Rearranging to find α:
\[ \alpha = \frac{25}{2 \times 10^{-2}} = \frac{25}{0.02} = 1250 \, \text{rad/s}^2 \]

### Final Answer
The angular acceleration of the ring at time \( t = 0 \) is:
\[ \alpha = 1250 \, \text{rad/s}^2 \]

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