A magnetic needle suspended parallel to the magnetic field requires √3 J of work to turn it through 60°. What will be the torque needed to maintain the needle in this position?(A) 3 J(B) √3 J(C) 3/2 J(D) 2√3 J

The work done in turning a magnetic needle through an angle in a magnetic field is given by the expression:

Work (W) = -M * B * cos(θ)

Where:

W is the work done.
M is the magnetic moment of the needle.
B is the magnetic field strength.
θ is the angle between the magnetic moment and the magnetic field.
In this case, we are given that the work done (W) is √3 J (square root of 3 joules) and the angle (θ) is 60°.

So, W = √3 J
θ = 60°

We want to find the torque (τ) needed to maintain the needle in this position. Torque is related to work by the formula:

τ = -W / θ

Substituting the given values:

τ = -(√3 J) / (60°)

Now, let's convert the angle from degrees to radians, as the trigonometric functions in the formula use radians:

1 radian = 180° / π
So, 60° = (60/180)π = (1/3)π radians

Now, plug this value back into the formula for torque:

τ = -(√3 J) / ((1/3)π radians)

To simplify, you can multiply the numerator and denominator by 3/π:

τ = -((3/π)√3 J)

Now, let's approximate the value of (3/π) to get the final answer:

τ ≈ - (0.9549√3 J)

The negative sign indicates that the torque is in the opposite direction of the magnetic field. So, the magnitude of the torque needed to maintain the needle in this position is approximately 0.9549√3 J.

The closest option to this magnitude is (D) 2√3 J.