Question

A bar magnet of length 10 m and magnetic moment 25 $Am^2$ is bent in the form of an arc as shown in figure. The new magnetic dipole moment is

• A. $\frac{75}{2}Am^2$
• B. $\frac{75\sqrt{3}}{2\pi}Am^2$
• C. $\frac{25}{2}Am^2$
• D. $\frac{25\sqrt{3}}{2\pi}Am^2$

Answer 1

Answer: (B)

$\frac{75\sqrt{3}}{2\pi}Am^2$

Answer 2

To solve this problem, we need to find the new effective length of the magnet after it's bent into an arc and then calculate the new magnetic dipole moment.

1. Understand the initial state:

   • Length of the bar magnet, $L = 10 \, \text{m}$.
   • Magnetic moment, $M = 25 \, \text{Am}^2$.
   • For a straight bar magnet, the magnetic moment is given by $M = m \times L$, where $m$ is the pole strength.
   • From this, we can find the pole strength: $m = \frac{M}{L} = \frac{25 \, \text{Am}^2}{10 \, \text{m}} = 2.5 \, \text{A m}$.
   • The pole strength ($m$) remains constant when the magnet is bent.

2. Understand the final state (bent magnet):

   • The magnet is bent into an arc subtending an angle $\theta = 120^\circ$ at the center.
   • Convert the angle to radians: $\theta = 120^\circ \times \frac{\pi}{180^\circ} = \frac{2\pi}{3}$ radians.
   • The length of the arc is the original length of the magnet, $L = 10 \, \text{m}$.
   • For an arc, the length $L = R\theta$, where $R$ is the radius of the circle of which the arc is a part.
   • We can find the radius $R$: $R = \frac{L}{\theta} = \frac{10 \, \text{m}}{2\pi/3} = \frac{30}{2\pi} = \frac{15}{\pi} \, \text{m}$.

3. Calculate the new effective length ($L'$):

   • The new magnetic dipole moment is determined by the straight-line distance between the two poles (ends of the arc). This distance is the chord length of the arc.
   • The chord length $L'$ for an arc subtending an angle $\theta$ at the center of a circle with radius $R$ is given by $L' = 2R \sin\left(\frac{\theta}{2}\right)$.
   • Calculate $\frac{\theta}{2}$: $\frac{\theta}{2} = \frac{120^\circ}{2} = 60^\circ$.
   • $L' = 2 \times \left(\frac{15}{\pi}\right) \times \sin(60^\circ)$.
   • We know $\sin(60^\circ) = \frac{\sqrt{3}}{2}$.
   • $L' = 2 \times \frac{15}{\pi} \times \frac{\sqrt{3}}{2} = \frac{15\sqrt{3}}{\pi} \, \text{m}$.

4. Calculate the new magnetic dipole moment ($M'$):

   • The new magnetic dipole moment $M' = m \times L'$.
   • $M' = 2.5 \, \text{A m} \times \frac{15\sqrt{3}}{\pi} \, \text{m}$.
   • $M' = \frac{5}{2} \times \frac{15\sqrt{3}}{\pi} = \frac{75\sqrt{3}}{2\pi} \, \text{Am}^2$.

The new magnetic dipole moment is $\frac{75\sqrt{3}}{2\pi} \, \text{Am}^2$.