Question

A magnet of length $0.1m$ and pole strength ${10^{ - 4}}A/m$ is kept in a magnetic field of $30Wb/{m^2}$ at an angle of ${30^0}$ . The couple acting on it is ${10^{ - 4}}Nm$ .

(A) $7.5$
(B) $3.0$
(C) $1.5$
(D) $6.0$

Answer 1

Hint:
First start with finding the relation between the couple or torque acting, magnetic dipole moment and the magnetic field and then put all the values from the question and get the required answer. Use the formula of dipole moment in terms of magnet length and pole strength.




Formula used :
torque acting on the magnet is given by;
$\tau = \vec M \times \vec B = MB\sin \theta $


Complete step by step solution:

From the question, we know that;
Length of the magnet is $0.1m$ .
Pole strength is ${10^{ - 4}}Am$.
Magnetic field is $30Wb/{m^2}$.

Now, we know that the torque acting on the magnet is given by;
$\tau = \vec M \times \vec B = MB\sin \theta $
Also, we know magnetic dipole moment in terms of length and pole strength:
$M = m \times 2l$
So, $\tau = m \times \left( {2l} \right) \times B\sin \theta $
$\tau = {10^{ - 4}} \times 0.1 \times 30\sin {30^0}$
By solving;
$\tau = 1.5 \times {10^{ - 4}}Nm$

Hence the correct answer is Option(C).







Note:
Use the formula of the magnetic dipole moment in terms of the magnet length and pole strength and put the value accordingly. Be careful about the unit of all the quantities before using in the formula used in the solution then only you will get the right answer if any of the units is not the same then answer will be different.