Question

The effective length of a magnet is $31.4\;cm\;$ and its pole strength is $0.8\;Am$. The magnetic moment, if it is bent in the form of a semicircle, is _____ $A{m^2}$.
A. $1.2$
B. $1.6$
C. $0.16$
D. $0.12$

Answer 1

Magnetic dipole moment is the measure of the product of the distance between the poles and the pole strength. It is also defined as the tendency of the magnet to get attracted to the magnetic field. This distance between the two poles present in the bar magnet is called the effective length of the magnet. This effective length of the magnet varies when the magnet changes its shape. Therefore we need to first find the magnetic moment formula for the semicircle and then substitute the given values to find the new magnetic moment.

Complete step by step solution:
Given that the bar magnet is bent to form a semi-circle. We know that the circumference of the semicircle is the length of the semi-circle which is said to be $l$ .
$l = \pi r$ …… (1)
Therefore the radius of the semi-circle will be,
$r = \dfrac{l}{\pi }$ ….. (2)
If the radius of the semi-circle is r then the distance between the poles will be $2r$. Therefore the above equation becomes,
$2r = \dfrac{{2l}}{\pi }$ …… (3)
Here l is the effective length of the bar magnet.
If we consider that this bar magnet has a magnetic moment M and pole strength m and the distance between the two poles is $2r$. Then the magnetic moment formula can be written as.
$M = m(2r)$
From equation (3), substitute in the above equation we get,
$M = m \times \dfrac{{2l}}{\pi }$ …… (4)
Therefore the above equation gives us the magnetic dipole moment formula for a semi-circle. Now we can substitute all the given values from the question in the above equation.
Given that the effective length of the magnet $l = 31.4\;cm\; = 31.4 \times {10^{ - 2}}m$
The pole strength of the magnet $m = 0.8\;Am$
Therefore we get,
$M = 0.8 \times \dfrac{{2 \times 31.4 \times {{10}^{ - 2}}}}{{3.14}}$
Solving the above equation we get,
$M = 0.16A{m^2}$

The net magnetic moment of the bar magnet which is bent to a semi-circle is given by $M = 0.16A{m^2}$.
Therefore the correct option is C.

Note:
There is indeed a difference between the effective length and the geometric length of the bar magnet. As we already said, the effective length of the bar magnet is said to be measured between the two poles of the bar magnet. The poles will not be located at the two edges of the bar magnet, they will be located a little inside the edges. Therefore the effective length of the bar magnet will be a little lesser than its geometric length.