Question

A magnetic rod with a length of 6.00 cm, radius 3.00 mm, and (uniform) magnetization \[2.70 \times {10^3}A/m\] can turn about its center like a compass needle. It is placed in a uniform magnetic field B of magnitude 35.0 mT, such that the directions of its dipole moment and make an angle of $68^\circ $.
(a) What is the magnitude of the torque on the rod due to B?
(b) What is the change in the orientation energy of the rod if the angle changes to $34^\circ $?

Answer 1

Hint: Magnetization is also called magnetic polarization and is the vector field that represents the density of the induced magnetic dipole moments in a material. It depends on the size of the dipole moments of the atoms in a given substance and also the degree to which these dipole moments are aligned to each other.

Complete step by step solution:
Given data:
The length of the rod, $L = 6cm = 6 \times {10^{ - 2}}m$
The radius of the rod, $R = 3mm = 3 \times {10^{ - 3}}m$
Magnetization, \[M = 2.70 \times {10^3}A/m\]
Initial angle $ = 68^\circ $
Final angle $ = 34^\circ $
Magnetic field, $B = 35mT = 35 \times {10^{ - 3}}T$
The magnitude of the torque on the rod due to B =?
The change in the orientation energy of the rod =?
a) We know that the magnetic moment of the rod is given by,
$\mu = M \times V$
$ \Rightarrow \mu = M \times \pi {R^2} \times L$
$ \Rightarrow \mu = 2.70 \times 1{0^3} \times \pi \times {\left( {3 \times {{10}^{ - 3}}} \right)^2} \times \left( {6 \times {{10}^{ - 2}}} \right)$
$ \Rightarrow \mu = 4.57 \times {10^{ - 3}}A{m^2}$
We know that the magnitude of the torque on the rod is given by,
$\tau = \mu B\sin \theta $
$ \Rightarrow \tau = \left( {4.57 \times {{10}^{ - 3}}} \right)\left( {35 \times {{10}^{ - 3}}} \right)\sin 68^\circ $
\[ \Rightarrow \tau = 1.48 \times {10^{ - 4}}Nm\]
Thus the magnitude of the torque on the rod due to B is \[1.48 \times {10^{ - 4}}Nm\].

b) Let $\Delta U$ be the change in the orientation energy of the rod.
Then the change in the orientation energy of the rod is given by,
$\Delta U = - \mu B\left( {\cos {\theta _f} - \cos {\theta _i}} \right)$
$ \Rightarrow \Delta U = - 4.57 \times {10^{ - 3}} \times 35 \times {10^{ - 3}}\left( {\cos 34^\circ - \cos 68^\circ } \right)$
$ \Rightarrow \Delta U = - 72.9\mu J$
Thus the change in the orientation energy of the rod $ = - 72.9\mu J$.

Note: 1. Torque is a twisting force and is defined as the measure of how the force that is acting on an object causes it to rotate. Torque is a vector quantity. Unit of torque is newton-meter. Torque depends on the direction of the force on the axis.
2. The magnetic dipole is a tiny magnet of the microscopic to the subatomic dimensions which is equal to the flow of the electric charge around the loop.