A magnetic needle suspended parallel to a magnetic field requires `sqrt3J` of work to turn it through `60^@`. The torque needed to maintain the needle in this postion will be:

Options

A. 3 J
B. `sqrt3J`
C. `(3)/(2)J`
D. `2sqrt3J`

Explanation

To solve the problem step by step, we need to determine the torque required to maintain a magnetic needle in a magnetic field after it has been turned through \(60^\circ\).

### Step 1: Understand the Work Done
We are given that the work done to turn the magnetic needle through \(60^\circ\) is \(\sqrt{3} \, J\). The work done in rotating a magnetic moment \(m\) in a magnetic field \(B\) is given by the formula:

\[ W = -mB (\cos \theta_2 - \cos \theta_1) \]

Where:
- \(W\) is the work done,
- \(m\) is the magnetic moment,
- \(B\) is the magnetic field strength,
- \(\theta_1\) is the initial angle (which is \(0^\circ\) since it starts parallel to the magnetic field),
- \(\theta_2\) is the final angle (\(60^\circ\)).

### Step 2: Substitute Values
Substituting the known values into the work formula:

\[ \sqrt{3} = -mB \left(\cos 60^\circ - \cos 0^\circ\right) \]

We know that:
- \(\cos 60^\circ = \frac{1}{2}\)
- \(\cos 0^\circ = 1\)

So, we can rewrite the equation as:

\[ \sqrt{3} = -mB \left(\frac{1}{2} - 1\right) \]

This simplifies to:

\[ \sqrt{3} = -mB \left(-\frac{1}{2}\right) = \frac{mB}{2} \]

### Step 3: Solve for \(mB\)
Rearranging gives us:

\[ mB = 2\sqrt{3} \]

### Step 4: Calculate the Torque
The torque (\(\tau\)) on a magnetic moment in a magnetic field is given by:

\[ \tau = mB \sin \theta \]

Where \(\theta\) is the angle between the magnetic moment and the magnetic field. In this case, after turning \(60^\circ\), we have:

\[ \tau = mB \sin 60^\circ \]

We know \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), so substituting \(mB\) into the torque equation gives:

\[ \tau = (2\sqrt{3}) \left(\frac{\sqrt{3}}{2}\right) \]

### Step 5: Simplify the Torque Expression
This simplifies to:

\[ \tau = 2\sqrt{3} \cdot \frac{\sqrt{3}}{2} = 3 \]

### Final Answer
Thus, the torque needed to maintain the needle in this position is:

\[ \tau = 3 \, \text{N m} \]

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