Question

A bar magnet placed in a uniform magnetic field making an angle $\theta$ with the field experiences a torque. If the angle made by the magnet with the field is doubled, the torque experienced by the magnet increases by $41.4\%$. The initial angle made by the magnet with the magnetic field is

• A. $60^{\circ}$
• B. $30^{\circ}$
• C. $90^{\circ}$
• D. $45^{\circ}$

Answer 1

The Correct Option is D

To solve this problem, we need to understand the relationship between the torque experienced by a bar magnet in a magnetic field and the angle it makes with the field.

The torque (\tau) experienced by a bar magnet in a uniform magnetic field is given by the formula:

\(\tau = MB \sin \theta\)

where M is the magnetic moment of the bar magnet, B is the magnetic field strength, and \theta is the angle between the magnetic moment and the magnetic field.

Initially, the angle is \theta, so the initial torque is:

\(\tau_1 = MB \sin \theta\)

When the angle is doubled, the new angle becomes 2\theta and the new torque is:

\(\tau_2 = MB \sin 2\theta\)

We know that \sin 2\theta = 2 \sin \theta \cos \theta, thus:

\(\tau_2 = MB \cdot 2 \sin \theta \cos \theta\)

The problem states that the torque increases by 41.4\% when the angle is doubled. Therefore, we can write:

\(\tau_2 = \tau_1 + 0.414 \tau_1\)

This simplifies to:

\(\tau_2 = 1.414 \tau_1\)

Substituting the expressions for \tau_1 and \tau_2:

MB \cdot 2 \sin \theta \cos \theta = 1.414 \times MB \cdot \sin \theta

Since MB is a common term, it can be canceled out from both sides:

2 \sin \theta \cos \theta = 1.414 \sin \theta

This simplifies to:

2 \cos \theta = 1.414

Solving for \cos \theta gives:

\(\cos \theta = \frac{1.414}{2} = 0.707\)

This value corresponds to an angle of 45^{\circ}. Therefore, the initial angle made by the magnet with the magnetic field is:

45^{\circ}