A coil having magnetic moment 15 A m² placed in a uniform magnetic field of 4 T in the horizontal direction exists such that initially the axis of coil is in the direction of the field. If the coil is rotated by `45^(@)` and the moment of inertia of the coil is `0.5" kg m"^(2)` then the angular speed acquired by the coil is

Options

A. `20 "rad s"^(-1)`
B. `10 "rad s"^(-1)`
C. `8.34 "rad s"^(-1)`
D. `4.5 "rad s"^(-1)`

Explanation

To solve the problem step by step, we need to calculate the angular speed acquired by the coil after it has been rotated by 45 degrees in a magnetic field. Here’s the detailed solution:

### Step 1: Understand the Problem
The coil has a magnetic moment \( M = 15 \, \text{A m}^2 \) and is placed in a magnetic field \( B = 4 \, \text{T} \). The coil is initially aligned with the magnetic field and is then rotated by \( 45^\circ \). The moment of inertia of the coil is given as \( I = 0.5 \, \text{kg m}^2 \).

### Step 2: Calculate the Torque
The torque \( \tau \) acting on the coil when it is rotated in a magnetic field is given by the formula:
\[ \tau = M B \sin \theta \]
where \( \theta \) is the angle between the magnetic moment and the magnetic field. After rotating the coil by \( 45^\circ \), we have:
\[ \theta = 45^\circ \]
Thus, we can calculate the torque:
\[ \tau = 15 \, \text{A m}^2 \times 4 \, \text{T} \times \sin(45^\circ) = 15 \times 4 \times \frac{1}{\sqrt{2}} = 30\sqrt{2} \, \text{N m} \]

### Step 3: Relate Torque to Angular Acceleration
The torque can also be expressed in terms of angular acceleration \( \alpha \):
\[ \tau = I \alpha \]
Since the coil is rotating, we can express \( \alpha \) in terms of angular velocity \( \omega \) and angular displacement \( \theta \):
\[ \tau = I \frac{d\omega}{dt} \]
We can relate \( \alpha \) to \( \omega \) by using the chain rule:
\[ \tau = I \omega \frac{d\omega}{d\theta} \]

### Step 4: Set Up the Equation
Equating the two expressions for torque:
\[ I \omega \frac{d\omega}{d\theta} = M B \sin \theta \]

### Step 5: Integrate the Equation
Rearranging gives:
\[ I \omega d\omega = M B \sin \theta d\theta \]
Integrating both sides:
\[ \int_0^\omega I \omega \, d\omega = \int_0^{45^\circ} M B \sin \theta \, d\theta \]
The left side becomes:
\[ \frac{I}{2} \omega^2 \]
The right side becomes:
\[ M B \left[-\cos \theta\right]_0^{45^\circ} = M B \left(-\cos(45^\circ) + \cos(0)\right) = M B \left(-\frac{1}{\sqrt{2}} + 1\right) \]

### Step 6: Substitute Values
Substituting \( M = 15 \), \( B = 4 \):
\[ \frac{I}{2} \omega^2 = 15 \times 4 \left(1 - \frac{1}{\sqrt{2}}\right) \]
Calculating the right side:
\[ = 60 \left(1 - \frac{1}{\sqrt{2}}\right) = 60 \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) = 60 \frac{\sqrt{2} - 1}{\sqrt{2}} \]

### Step 7: Solve for Angular Speed \( \omega \)
Now substituting \( I = 0.5 \):
\[ \frac{0.5}{2} \omega^2 = 60 \frac{\sqrt{2} - 1}{\sqrt{2}} \]
\[ 0.25 \omega^2 = 60 \frac{\sqrt{2} - 1}{\sqrt{2}} \]
\[ \omega^2 = \frac{60 \cdot 4(\sqrt{2} - 1)}{1} = 240(\sqrt{2} - 1) \]
\[ \omega = \sqrt{240(\sqrt{2} - 1)} \]
Calculating this gives approximately \( \omega \approx 8.34 \, \text{rad/s} \).

### Final Answer
The angular speed acquired by the coil is approximately \( 8.34 \, \text{rad/s} \).

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