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 solve the problem, we will use the principle of conservation of energy. The total mechanical energy (kinetic + potential) remains constant in the absence of non-conservative forces.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Moment of inertia of the coil, \( I = 0.8 \, \text{kg m}^2 \)
- Magnetic field strength, \( B = 4 \, \text{T} \)
- Angle between magnetic field and magnetic moment, \( \theta = 60^\circ \)
- Magnetic moment of the coil, \( m = 20 \, \text{A m}^2 \)

2. **Calculate Initial Potential Energy**:
The potential energy \( U \) in a magnetic field is given by:
\[ U = -\vec{m} \cdot \vec{B} = -mB \cos \theta \]
Substituting the values:
\[ U_{\text{initial}} = -20 \times 4 \times \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ U_{\text{initial}} = -20 \times 4 \times \frac{1}{2} = -40 \, \text{J} \]

3. **Calculate Final Potential Energy**:
When the coil passes through stable equilibrium, the angle becomes \( 0^\circ \) (aligned with the magnetic field), so:
\[ U_{\text{final}} = -mB \cos(0^\circ) = -mB \cdot 1 = -20 \times 4 \times 1 = -80 \, \text{J} \]

4. **Apply Conservation of Energy**:
The total mechanical energy is conserved:
\[ K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}} \]
Here, the initial kinetic energy \( K_{\text{initial}} = 0 \) (as it is released from rest), so:
\[ 0 + (-40) = K_{\text{final}} + (-80) \]
Rearranging gives:
\[ K_{\text{final}} = -40 + 80 = 40 \, \text{J} \]

5. **Express Final Kinetic Energy**:
The final kinetic energy is given by:
\[ K_{\text{final}} = \frac{1}{2} I \omega^2 \]
Setting this equal to the kinetic energy we found:
\[ \frac{1}{2} I \omega^2 = 40 \]
Substituting \( I = 0.8 \):
\[ \frac{1}{2} \times 0.8 \times \omega^2 = 40 \]
\[ 0.4 \omega^2 = 40 \]
\[ \omega^2 = \frac{40}{0.4} = 100 \]
\[ \omega = \sqrt{100} = 10 \, \text{rad/s} \]

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

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