A bar magnet having a magnetic moment of \(2 \times 10^{4} \mathrm{JT}^{-1}\) is free to rotate in a horizontal plane. A horizontal magnetic field \(\mathrm{B}=6 \times 10^{-4}\) Tesla exists in the space. The work done in taking the magnet slowly from a direction parallel to the field to a direction \(60^{\circ}\) from the field is (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\)

We are given the following information: - The bar magnet's magnetic moment: \(m = 2 \times 10^4 \, \mathrm{JT^{-1}}\) - The horizontal magnetic field: \(B = 6 \times 10^{-4} \, \mathrm{T}\) (Tesla) - The angle the magnet is rotated: \(\theta = 60^{\circ}\) We need to find the work done to rotate the magnet.

The expression for the work done, \(W\), in rotating a magnet in a magnetic field is given by: \[ W = -mB(\cos{\theta_1}-\cos{\theta_2})\] Where, - \(W\) is the work done, - \(m\) is the magnetic moment of the magnet, - \(B\) is the magnetic field, - \(\theta_1\) is the initial angle between the magnetic moment and the magnetic field, - \(\theta_2\) is the final angle between the magnetic moment and the magnetic field. In this problem, the magnet is initially parallel to the field (\(\theta_1 = 0^{\circ}\)) and is finally at an angle of \(60^{\circ}\) (\(\theta_2 = 60^{\circ}\)) from the field.

Now, we can plug in the given values into the formula for the work done: \[W = -(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})(\cos{0^{\circ}} - \cos{60^{\circ}})\] To calculate the value, we must find the cosine of the angles. Remember that \(\cos{0^{\circ}} =1\) and \(\cos{60^{\circ}} = \frac{1}{2}\). \[W = -(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})(1 - \frac{1}{2})\] Now, we can perform the calculations: \[W = -(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})(\frac{1}{2})\] \[W = -\frac{1}{2}(2 \times 10^4\,\mathrm{JT^{-1}})(6\times10^{-4}\,\mathrm{T})\] \[W = -\frac{1}{2}(12\,\mathrm{J})\] \[W = -6\,\mathrm{J}\] Here, the negative sign indicates that work has been done on the magnet to rotate it. The absolute value of the work done is \(6 \, \mathrm{J}\).

Compare the calculated work done to the given options: (a) \(0.6 \mathrm{~J}\) (b) \(12 \mathrm{~J}\) (c) \(6 \mathrm{~J}\) (d) \(2 \mathrm{~J}\) Option (c) matches our calculated work done of \(6 \, \mathrm{J}\). Therefore, the correct answer is (c) \(6 \mathrm{~J}\).