Question

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

• A. 80 J
• B. \( -40\sqrt{3} \) J
• C. -60 J
• D. -80 J

Hint:

Remember the formulas for the torque \( \tau = MB \sin \theta \) and potential energy \( U = -MB \cos \theta \) of a magnetic dipole in a uniform magnetic field. Use the given torque and angle to find the product \( MB \), and then use this value to calculate the potential energy at the same angle.

Answer 1

The Correct Option is D

The torque \( \tau \) experienced by a magnetic dipole in a uniform magnetic field \( \vec{B} \) is given by: \[ \tau = \vec{M} \times \vec{B} = MB \sin \theta \] where \( M \) is the magnitude of the magnetic dipole moment, \( B \) is the magnitude of the magnetic field, and \( \theta \) is the angle between \( \vec{M} \) and \( \vec{B} \). Given \( \tau = 80\sqrt{3} \) N m and \( \theta = 60^\circ \). \[ 80\sqrt{3} = MB \sin 60^\circ \] \[ 80\sqrt{3} = MB \left( \frac{\sqrt{3}}{2} \right) \] \[ MB = \frac{80\sqrt{3} \times 2}{\sqrt{3}} = 160 \] The potential energy \( U \) of the magnetic dipole in the uniform magnetic field is given by: \[ U = -\vec{M} \cdot \vec{B} = -MB \cos \theta \] Substituting the values \( MB = 160 \) and \( \theta = 60^\circ \): \[ U = -(160) \cos 60^\circ \] \[ U = -160 \left( \frac{1}{2} \right) \] \[ U = -80 \text{ J} \]

Answer 2

To find the potential energy of a magnetic dipole in a uniform magnetic field, we need to understand the relationship between torque, magnetic dipole moment, and the angle with the magnetic field. Given data includes:

• Torque \( \tau = 80\sqrt{3} \) Nm
• Angle \( \theta = 60^\circ \)

The torque experienced by a magnetic dipole in a magnetic field is given by:

\[\tau = mB\sin\theta\]

where:

• \(m\) = magnetic dipole moment
• \(B\) = magnetic field strength
• \(\theta\) = angle between magnetic dipole and magnetic field

We are interested in finding the potential energy (\(U\)) of the dipole, which is given by:

\[U = -mB\cos\theta\]

First, solve for \(mB\) using the torque formula:

\[80\sqrt{3} = mB\sin60^\circ = mB\cdot \frac{\sqrt{3}}{2}\]

Simplify to find:

\[mB = \frac{80\sqrt{3}}{\sqrt{3}/2} = 160\]

Now, substitute \(mB\) in the potential energy formula:

\[U = -160\cos60^\circ\]

Since \(\cos60^\circ = \frac{1}{2}\), we find:

\[U = -160 \times \frac{1}{2} = -80 \text{ J}\]

Thus, the potential energy of the dipole in the magnetic field is -80 J.

This matches the correct option provided: -80 J.