Question 9: Magnetism and Matter

Question

A magnetic dipole experiences a torque of $80 \sqrt{3} \mathrm{~N} \mathrm{~m}$ when placed in uniform magnetic field in such a way that dipole moment makes angle of $60^{\circ}$ with magnetic field. The potential energy of the dipole is :

Accepted Answer

To determine the potential energy of a magnetic dipole in a uniform magnetic field, we start with the torque equation:
$ 	au = M 	imes B = MB \sin 60^\circ $
Given that $	au = 80 \sqrt{3} \, 	ext{N m}$ and $\sin 60^\circ = \frac{\sqrt{3}}{2}$, we can equate and solve for $MB$:
$ MB \frac{\sqrt{3}}{2} = 80 \sqrt{3} $
Solving for $MB$, we find:
$ MB = 160 $
The potential energy $U$ of the dipole in the magnetic field is given by:
$ U = -M \cdot B = -MB \cos 60^\circ $
Substituting in $\cos 60^\circ = \frac{1}{2}$, we calculate:
$ U = -160 	imes \frac{1}{2} = -80 \, 	ext{J} $
Thus, the potential energy of the dipole is $-80 \, 	ext{J}$.

Answer

$$ 	ext { }-40 \sqrt{3} \mathrm{~J} $$

Answer

-80 J

Answer

80 J

Answer

-60 J

Magnetism and Matter

A magnetic dipole experiences a torque of 80 \sqrt{3} \mathrm{~N} \mathrm{~m} when placed in uniform magnetic field in such a way that dipole moment makes angle of 60^{\circ} with magnetic field. The potential energy of the dipole is :

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