Magnetic Moment Calculation for Bent Bar
The magnetic moment ($M$) of a bar magnet is typically defined as the product of its pole strength ($q_m$) and its length ($L$), represented vectorially as $\vec{M} = q_m \vec{L}$. We assume the magnetic moment acts along the length of the bar.
Calculating New Magnetic Moment
The original bar has magnetic moment $M$. Let its length be $L$. So, $M = q_m L$.
When the bar is bent into an arc of a circle subtending an angle $\theta = 60^{\circ}$ at the center, the arc length is equal to the original length $L$. The angle in radians is $\theta = 60^{\circ} = \frac{60}{180} \pi = \frac{\pi}{3}$ radians.
Let $r$ be the radius of the circular arc. The arc length is given by $L = r \theta$.
Therefore, $L = r \left(\frac{\pi}{3}\right)$.
This gives the radius as $r = \frac{3L}{\pi}$.
The new magnetic moment, $M'$, is associated with the straight-line distance (chord length, $C$) between the two ends of the bent arc. The magnetic moment vector is considered along this chord.
So, $M' = q_m C$.
The chord length $C$ is calculated using the formula $C = 2r \sin(\frac{\theta}{2})$.
Substituting the values:
$C = 2r \sin\left(\frac{60^{\circ}}{2}\right) = 2r \sin(30^{\circ})$
Since $\sin(30^{\circ}) = \frac{1}{2}$,
$C = 2r \left(\frac{1}{2}\right) = r$.
Now substitute the expression for $r$ back into the equation for $M'$:
$M' = q_m C = q_m r = q_m \left(\frac{3L}{\pi}\right)$
$M' = \frac{3}{\pi} (q_m L)$
Since the original magnetic moment $M = q_m L$, the new magnetic moment is:
$M' = \frac{3M}{\pi}$
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