Question

\(\mathrm{A}\) bar magnet of length \(10 \mathrm{~cm}\) and having the pole strength equal \(10^{3} \mathrm{Am}\) to is kept in a magnetic field having magnetic induction (B) equal to \(4 \pi \times 10^{3}\) tesla. It makes an angle of \(30^{\circ}\) with the direction of magnetic induction. The value of the torque acting on the magnet is Joule. (a) \(2 \pi \times 10^{-7}\) (b) \(2 \pi \times 10^{5}\) (c) \(0.5\) (d) \(0.5 \times 10^{2}\)

Short answer

The torque acting on the magnet is \(2\pi \times 10^{5} J\).

Step-by-step solution

Identify the formula for torque.

We know that the formula for torque acting on a magnet in a magnetic field is given by: \[\tau = pB \sin{\theta}\] Where: - \(\tau\) is the torque acting on the magnet - \(p\) is the magnetic moment of the magnet - \(B\) is the magnetic induction - \(\theta\) is the angle between the magnet and the magnetic induction direction. Since pole strength and length are given, we need to find the magnetic moment of the magnet first. The magnetic moment is calculated by the product of pole strength and length: \[p = m\times l\]

Find the magnetic moment of the magnet.

We are given the pole strength as \(10^3 Am\) and the length as \(10cm = 0.1m\). Thus, the magnetic moment is: \[p = (10^3 Am) \times (0.1 m) = 100 Am^2\]

Plug the values into the torque formula.

Now that we have the magnetic moment, we can plug the values into the torque formula: \[\tau = pB \sin{\theta} = (100 Am^2)(4\pi\times10^3 T)\sin{30^{\circ}}\]

Calculate the torque.

Now we need to calculate the value of the torque: \[\tau = (100 Am^2)(4\pi\times10^3 T)\sin{30^{\circ}} = 100 \times 4 \pi \times 10^3 \times\frac{1}{2} \] \[\tau = 2\pi \times 10^5Nm\]

Determine the torque in Joules.

We can express the torque value in Joule as well: \[1N.m = 1J\] Therefore, \[\tau = 2\pi \times 10^5J\] The value of the torque acting on the magnet is \(2\pi \times 10^{5} J\). Thus, the correct answer is (b) \(2 \pi \times 10^{5}\).