A magnetic needle lying parallel to the magnetic field required W units of work to turn it through an angle `45^@` The torque required to maintain the needle in this position will be

Options

A. `sqrt2W`
B. `1/(sqrt(3W))`
C. `(sqrt2-1)W`
D. `W/(sqrt(2)-1)`

Answer

To solve the problem, we need to find the torque required to maintain a magnetic needle in a position after it has been turned through an angle of \(45^\circ\) in a magnetic field. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Work Done
The work done \(W\) to turn the magnetic needle through an angle of \(45^\circ\) can be related to the change in potential energy of the magnetic dipole in the magnetic field. The potential energy \(U\) of a magnetic dipole in a magnetic field is given by:
\[ U = -\vec{M} \cdot \vec{B} = -MB \cos \theta \]
where \(M\) is the magnetic moment, \(B\) is the magnetic field strength, and \(\theta\) is the angle between the magnetic moment and the magnetic field.

### Step 2: Calculate Initial and Final Energies
1. **Initial Energy (\(U_i\))**: When the needle is parallel to the magnetic field (\(\theta = 0^\circ\)):
\[ U_i = -MB \cos(0) = -MB \]

2. **Final Energy (\(U_f\))**: When the needle is turned to \(45^\circ\):
\[ U_f = -MB \cos(45^\circ) = -MB \cdot \frac{1}{\sqrt{2}} = -\frac{MB}{\sqrt{2}} \]

### Step 3: Calculate Work Done
The work done \(W\) in turning the needle from \(0^\circ\) to \(45^\circ\) is given by the change in potential energy:
\[ W = U_f - U_i = \left(-\frac{MB}{\sqrt{2}}\right) - \left(-MB\right) \]
\[ W = MB - \frac{MB}{\sqrt{2}} = MB \left(1 - \frac{1}{\sqrt{2}}\right) \]

### Step 4: Relate Work Done to Magnetic Moment
From the expression for work done, we can express \(MB\) in terms of \(W\):
\[ MB = \frac{W}{1 - \frac{1}{\sqrt{2}}} \]

### Step 5: Calculate Torque
The torque \(\tau\) required to maintain the needle at \(45^\circ\) is given by:
\[ \tau = MB \sin(45^\circ) = MB \cdot \frac{1}{\sqrt{2}} \]
Substituting the expression for \(MB\) we derived:
\[ \tau = \left(\frac{W}{1 - \frac{1}{\sqrt{2}}}\right) \cdot \frac{1}{\sqrt{2}} \]
\[ \tau = \frac{W}{\sqrt{2} \left(1 - \frac{1}{\sqrt{2}}\right)} \]

### Step 6: Simplify the Expression
To simplify:
\[ \tau = \frac{W}{\sqrt{2} \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right)} = \frac{W \cdot \sqrt{2}}{\sqrt{2} - 1} \]

### Final Answer
The torque required to maintain the needle in the \(45^\circ\) position is:
\[ \tau = \frac{W \cdot \sqrt{2}}{\sqrt{2} - 1} \]