A current carrying ring with its center at origin and moment of inertia `10^(-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.
Maximum angular velocity of the ring in `rad//s` will be

Options

A. `50sqrt2`
B. `25sqrt2`
C. `100sqrt2`
D. `150sqrt2`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have a current-carrying ring with:
- Moment of Inertia, \( I = 10^{-2} \, \text{kg m}^2 \)
- Magnetic Moment, \( \mathbf{M} = (3\hat{i} - 4\hat{j}) \, \text{A m}^2 \)
- Magnetic Field, \( \mathbf{B} = (4\hat{i} + 3\hat{j}) \, \text{T} \)

### Step 2: Calculate the magnitudes of the magnetic moment and magnetic field
The magnitude of the magnetic moment \( M \) can be calculated as:
\[ M = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \, \text{A m}^2 \]

The magnitude of the magnetic field \( B \) can be calculated as:
\[ B = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \, \text{T} \]

### Step 3: Calculate the potential energy
The potential energy \( U \) in the magnetic field is given by:
\[ U = -\mathbf{M} \cdot \mathbf{B} = -MB \cos \theta \]
To find the maximum potential energy, we consider \( \theta = 180^\circ \) (which gives \( \cos(180^\circ) = -1 \)):
\[ U_{\text{max}} = MB = 5 \times 5 = 25 \, \text{J} \]

### Step 4: Relate potential energy to kinetic energy
According to the conservation of energy, the loss in potential energy will equal the gain in rotational kinetic energy:
\[ \Delta U = K.E. \]
The rotational kinetic energy \( K.E. \) is given by:
\[ K.E. = \frac{1}{2} I \omega^2 \]
Setting the potential energy equal to the kinetic energy:
\[ 25 = \frac{1}{2} (10^{-2}) \omega^2 \]

### Step 5: Solve for angular velocity \( \omega \)
Rearranging the equation gives:
\[ 25 = 0.005 \omega^2 \]
\[ \omega^2 = \frac{25}{0.005} = 5000 \]
\[ \omega = \sqrt{5000} = 50\sqrt{2} \, \text{rad/s} \]

### Final Answer
The maximum angular velocity of the ring is:
\[ \omega_{\text{max}} = 50\sqrt{2} \, \text{rad/s} \]