Question:
A rectangular loop of wire, supporting a mass \(m\), hangs with one end in a uniform magnetic field \(\vec{B}\) pointing out of the plane of the paper. A clockwise current is set up such that \(i \gt mg/Ba\), where \(a\) is the width of the loop. 
A rectangular loop of wire, supporting a mass \(m\), hangs with one end in a uniform magnetic field \(\vec{B}\) pointing out of the plane of the paper. A clockwise current is set up such that \(i \gt mg/Ba\), where \(a\) is the width of the loop.
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The vertical force on the top arm in the field is \(iBa\gt mg\); the magnetic force itself does no work.
Updated On: Oct 31, 2026
- The weight rises due to a vertical force caused by the magnetic field and work is done on the system.
- The weight do not rise due to vertical for caused by the magnetic field and work is done on the system.
- The weight rises due to a vertical force caused by the magnetic field but no work is done on the system.
- The weight rises due to a vertical force caused by the magnetic field and work is extracted from the magnetic field.
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The Correct Option is C
Solution and Explanation
Only the top arm PQ (length \(a\)) is in the field; the horizontal forces on the vertical arms cancel. The force on PQ is vertical with magnitude \(F=iBa\).
Given \(i\gt mg/(Ba)\), so \(F=iBa\gt mg\): the net vertical force is upward and the weight rises.
The magnetic force on moving charges is always perpendicular to their velocity (\(\vec F=q\vec v\times\vec B\)), so the magnetic field does no work and no energy is extracted from it; the energy for lifting the weight comes from the source maintaining the current.
So the weight rises due to a vertical force of magnetic origin, but the magnetic field itself does no work. This matches option (3).
Answer: (3) The weight rises due to a vertical force caused by the magnetic field but no work is done (by the magnetic field)
Given \(i\gt mg/(Ba)\), so \(F=iBa\gt mg\): the net vertical force is upward and the weight rises.
The magnetic force on moving charges is always perpendicular to their velocity (\(\vec F=q\vec v\times\vec B\)), so the magnetic field does no work and no energy is extracted from it; the energy for lifting the weight comes from the source maintaining the current.
So the weight rises due to a vertical force of magnetic origin, but the magnetic field itself does no work. This matches option (3).
Answer: (3) The weight rises due to a vertical force caused by the magnetic field but no work is done (by the magnetic field)
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