Question

A bar magnet when placed at an angle of \[30^\circ \] to the direction of magnetic field of induction of \[5 \times {\text{ }}{10^{ - 5}}T\], experiences a moment of a couple \[2.5 \times {\text{ }}{10^{ - 6}}N - m\], If the length of the magnet is \[5cm\] its pole strength is
A.\[2 \times {\text{ }}{10^{ - 2}}Am\]
B.\[5 \times {\text{ }}{10^{ - 2}}Am\]
C.\[2Am\]
D.\[5Am\]

Answer 1

Hint: Torque is defined as a coupled force acting on a body that causes a body to rotate about its axis.
Example: While opening a lead we apply coupled force (parallel, equal but in opposite direction ) to rotate it about its center.
Pole strength is a scalar quantity and it is defined as the strength of a bar magnet and its ability to attract other magnetic material towards itself and magnetic moment of a bar magnet is directly proportional to pole strength of a magnet.
Formula Used:
The torque acting on a bar magnet, \[\tau {\text{ }} = {\text{ }}MB{\text{ }}sin{\text{ }}\theta \]
Pole strength of bar magnet, $m = \dfrac{M}{L}$

Complete answer:
Given that,
The angle of rotation, \[\theta {\text{ }} = {\text{ }}30^\circ \]
magnetic field of induction, \[B{\text{ }} = {\text{ }}5 \times {\text{ }}{10^{ - 5}}T\]
Torque, \[\tau {\text{ }} = {\text{ }}2.5 \times {\text{ }}{10^{ - 6}}N - m\]
Length of the magnet, \[L = 5cm = 0.05m\]
The net torque acting on a bar magnet is given as
\[\tau {\text{ }} = {\text{ }}MB{\text{ }}sin{\text{ }}\theta \]
Substituting the given values of τ, B, and θ we get
\[2.5 \times {\text{ }}{10^{ - 6}} = {\text{ }}M{\text{ }} \times {\text{ }}5 \times {\text{ }}{10^{ - 5}} \times {\text{ }}sin{\text{ }}30\]
$M = \dfrac{{2.5 \times {\text{}}{{10}^{ - 6}}}}{{sin30 \times 5 \times {\text{}}{{10}^{ - 5}}}}$
$M = \dfrac{{2.5 \times {\text{}}{{10}^{ - 6}}}}{{0.5 \times 5 \times {\text{}}{{10}^{ - 5}}}}$
\[\therefore M{\text{ }} = 0.1{\text{ }}A{\text{ }}{m^2}\]
Now the pole strength of a magnet is given as
\[M = mL\]
\[0.1{\text{ }} = m \times {\text{ }}0.05\]
$\therefore m = \dfrac{{0.1}}{{0.05}} = 2 \times {10^{ - 3}} = 2mA$

Option C is correct among all.

Note:
In physics, torque is considered as a rotational analog of force, and it is expressed as $\vec \tau = \vec r \times \vec F$.
Also, we must note that if we cut a bar magnet in half its pole strength will remain unaffected whereas its magnetic dipole moment will become half of its original intensity.