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. (a)`sqrt2W`
B. (b)`1/(sqrt(3W))`
C. (c)`(sqrt2-1)W`
D. (d)`W/(sqrt(2)-1)`

Answer

To solve the problem step by step, we will analyze the work done in turning a magnetic needle and relate it to the torque required to maintain it in the new position.

### Step 1: Understand the Work Done
The work done \( W \) to turn a magnetic needle from an angle \( \theta_1 = 0^\circ \) (parallel to the magnetic field) to \( \theta_2 = 45^\circ \) can be expressed using the formula:
\[ W = M \cdot B \cdot (\cos \theta_1 - \cos \theta_2) \]
where:
- \( M \) is the magnetic moment,
- \( B \) is the magnetic field strength,
- \( \theta_1 \) and \( \theta_2 \) are the initial and final angles respectively.

### Step 2: Substitute the Angles
Substituting the values of \( \theta_1 \) and \( \theta_2 \):
\[ W = M \cdot B \cdot (\cos 0^\circ - \cos 45^\circ) \]
We know that:
- \( \cos 0^\circ = 1 \)
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)

Thus, we have:
\[ W = M \cdot B \cdot \left(1 - \frac{1}{\sqrt{2}}\right) \]

### Step 3: Simplify the Expression
Now, simplifying the expression:
\[ W = M \cdot B \cdot \left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]

### Step 4: Relate Torque to Work Done
The torque \( \tau \) required to maintain the needle at an angle \( \theta = 45^\circ \) can be expressed as:
\[ \tau = M \cdot B \cdot \sin \theta \]
Substituting \( \theta = 45^\circ \):
\[ \tau = M \cdot B \cdot \sin 45^\circ \]
Since \( \sin 45^\circ = \frac{1}{\sqrt{2}} \), we have:
\[ \tau = M \cdot B \cdot \frac{1}{\sqrt{2}} \]

### Step 5: Relate Torque to Work Done
From our earlier expression for work done, we can express \( M \cdot B \):
\[ M \cdot B = \frac{W \cdot \sqrt{2}}{\sqrt{2} - 1} \]
Now substituting this into the torque equation:
\[ \tau = \left(\frac{W \cdot \sqrt{2}}{\sqrt{2} - 1}\right) \cdot \frac{1}{\sqrt{2}} \]
This simplifies to:
\[ \tau = \frac{W}{\sqrt{2} - 1} \]

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