A coil has moment of inertia `0.8 kg/m^2` released in uniform magnetic field `4T` when there is `60^@` angle between magnetic field and magnetic moment of coil. Magnetic moment of coil is `20 A-m^2`. Find the angular speed of coil when it passes through stable equilibrium.

Options

A. `20 pi rad/S^(-1)`
B. `20 rad/S^(-1)`
C. `10 pi rad/S^(-1)`
D. `10 rad/s^(-1)`

Answer

To find the angular speed of the coil when it passes through stable equilibrium, we can follow these steps:

### Step 1: Understand the Initial and Final States
The coil is released from an angle of \(60^\circ\) with respect to the magnetic field. At this position, it has potential energy due to its orientation in the magnetic field. When it passes through stable equilibrium, the angle becomes \(0^\circ\), and the potential energy is minimized.

### Step 2: Write the Expression for Potential Energy
The potential energy \(U\) of the coil in a magnetic field is given by:
\[ U = -\mathbf{M} \cdot \mathbf{B} = -MB \cos \theta \]
where:
- \(M\) is the magnetic moment of the coil,
- \(B\) is the magnetic field strength,
- \(\theta\) is the angle between the magnetic moment and the magnetic field.

### Step 3: Calculate Initial Potential Energy
Initially, when \(\theta = 60^\circ\):
\[ U_i = -MB \cos(60^\circ) = -MB \cdot \frac{1}{2} \]
Substituting \(M = 20 \, \text{A-m}^2\) and \(B = 4 \, \text{T}\):
\[ U_i = -20 \cdot 4 \cdot \frac{1}{2} = -40 \, \text{J} \]

### Step 4: Calculate Final Potential Energy
When the coil passes through stable equilibrium, \(\theta = 0^\circ\):
\[ U_f = -MB \cos(0^\circ) = -MB \]
Substituting the values:
\[ U_f = -20 \cdot 4 = -80 \, \text{J} \]

### Step 5: Apply Conservation of Mechanical Energy
According to the conservation of mechanical energy:
\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]
Since the coil is released from rest, the initial kinetic energy is \(0\):
\[ 0 + U_i = \frac{1}{2} I \omega^2 + U_f \]
Substituting the potential energies:
\[ -40 = \frac{1}{2} I \omega^2 - 80 \]

### Step 6: Rearranging the Equation
Rearranging gives:
\[ \frac{1}{2} I \omega^2 = -40 + 80 \]
\[ \frac{1}{2} I \omega^2 = 40 \]

### Step 7: Solve for Angular Speed \(\omega\)
Multiply both sides by \(2\):
\[ I \omega^2 = 80 \]
Now, substituting the moment of inertia \(I = 0.8 \, \text{kg/m}^2\):
\[ 0.8 \omega^2 = 80 \]
Dividing both sides by \(0.8\):
\[ \omega^2 = \frac{80}{0.8} = 100 \]
Taking the square root:
\[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

### Final Answer
The angular speed of the coil when it passes through stable equilibrium is:
\[ \omega = 10 \, \text{rad/s} \]