Force On A Curved Current-carrying Wire
A semicircular wire of radius R carrying current I is placed in a uniform magnetic field B directed perpendicular to the plane of the semicircle. What is the magnitude of the net magnetic force on the curved semicircular portion?
Select the correct option:
Solution
2BIR
The magnetic force on a current-carrying conductor in a uniform field is F=I∫dl×B. A powerful simplification applies for a uniform field: the net force on any curved wire equals the force on a straight wire joining its endpoints, because ∫dl reduces to the straight-line vector L between the ends. For a semicircle of radius R, the two endpoints of the curved arc are separated by the diameter, so the effective length is L=2R. With the field perpendicular to the plane containing the wire, the angle between this effective length vector and B is 90∘, giving F=BI(2R)=2BIR. The option πBIR wrongly uses the full arc length πR, which would only matter if the field varied along the wire. The option BIR uses R instead of the diameter 2R. The option 0 would hold for a complete closed loop, where the endpoints coincide, but a semicircle is open. This effective-length theorem is a standard NCERT-level result. A check confirms that the force scales with the straight-line separation of the endpoints, here the diameter. This effective-length theorem dramatically simplifies otherwise tedious integrals and is well worth memorising for examination problems involving arcs, zigzags, and irregular bends in uniform fields. Its deeper meaning is that a uniform field exerts no net force on a closed current loop, since the closing endpoints coincide and the effective length vanishes; the loop instead experiences only a torque, which is exactly why current loops behave as magnetic dipoles that rotate rather than translate.
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About This Question
- Subject
- physics
- Chapter
- magnetic effects of current and magnetism
- Topic
- force on a curved current-carrying wire
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2BIR
The magnetic force on a current-carrying conductor in a uniform field is F=I∫dl×B. A powerful simplification applies for a uniform field: the net force on any curved wire equals the force on a straight wire joining its endpoints, because ∫dl reduces to the straight-line vector L between the ends. For a semicircle of radius R, the two endpoints of the curved arc are separated by the diameter, so the effective length is L=2R. With the field perpendicular to the plane containing the wire, the angle between this effective length vector and B is 90∘, giving F=BI(2R)=2BIR. The option πBIR wrongly uses the full arc length πR, which would only matter if the field varied along the wire. The option BIR uses R instead of the diameter 2R. The option 0 would hold for a complete closed loop, where the endpoints coincide, but a semicircle is open. This effective-length theorem is a standard NCERT-level result. A check confirms that the force scales with the straight-line separation of the endpoints, here the diameter. This effective-length theorem dramatically simplifies otherwise tedious integrals and is well worth memorising for examination problems involving arcs, zigzags, and irregular bends in uniform fields. Its deeper meaning is that a uniform field exerts no net force on a closed current loop, since the closing endpoints coincide and the effective length vanishes; the loop instead experiences only a torque, which is exactly why current loops behave as magnetic dipoles that rotate rather than translate.
This hard difficulty physics question is from the chapter magnetic effects of current and magnetism, covering the topic of force on a curved current-carrying wire. It appeared in the 2025 exam.
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