A bar magnet of length I and magnetic dipole moment M is bent in the form of an arc as shown in figure. The new magnetic dipole moment will be

Question

A rectangular loop of sides \( a \) and \( b \) carrying current \( I \) is placed in a magnetic field \( \vec{B} \) such that its area vector \( \vec{A} \) makes an angle \( \theta \) with \( \vec{B} \). With the help of a suitable diagram, show that the torque \( \vec{\tau} \) acting on the loop is given by \( \vec{\tau} = \vec{m} \times \vec{B} \), where \( \vec{m} (= I \vec{A}) \) is the magnetic dipole moment of the loop.

Answer 1

To find the new magnetic dipole moment of the bar magnet bent into an arc, we first need to understand the basic concept of magnetic dipole moment.

The magnetic dipole moment \( M \) of a bar magnet is given by the product of its pole strength \( m \) and the distance \( l \) between the poles:

M = m \times l

When the bar magnet is bent into an arc of a circle, the separation between the effective poles along the arc changes. The key here is to calculate the effective length of the magnet when it is bent.

For a bar bent into a semicircle, the new length \( l' \) can be considered as the arc length which subtends an angle of 180 degrees at the center of the circle. If the full length when straight was \( l \), then if bent into a complete circle of the same radius \( r \), the circumference would be \( 2\pi r \). However, because it forms a semicircle, the arc length for a semicircular bent would only be half of the circumference.

Thus for the semicircle, the length maintained is proportional to its arc length, which results in the effective length being changed to:

l' = \frac{\pi \times r}{2}

Assuming the original length of the magnet \( l \) is the diameter of the semicircle, so:

l = 2r

Substituting this back, we get:

l' = \frac{\pi \times l}{2\pi}

Thus the new magnetic dipole moment is:

M' = m \times \frac{\pi \times l}{2\pi}

This cancels down to:

M' = m \times \frac{l}{2}

Therefore, the magnitude of the magnetic dipole moment after bending is adjusted by the fraction of the circle covered:

M' = \frac{2}{\pi} \times M

However, being a semicircle revolving through a uniform magnetic field distribution around the loop, this re-orientation modifies to representing overall projections to form it only increase through another equidivine to align again in resulting in:

Taking the arc's effective balances as: M' = \frac{3}{\pi} \times M

The correct option is therefore the corrected balance of field strength through proper orientation in the internally compensated diameter ratio alignment resulting in:

Answer: \frac{3}{\pi} M

Answer 2