A wire is bent as shown in figure is placed in a uniform field B that emerges from the plane of the paper, outward as shown. The magnitude of force on the given wire is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4BIR \)
In a uniform magnetic field, the force on a current-carrying wire of any shape equals the force on a straight wire joining its two ends, carrying the same current. This is because \( \vec{F} = I\int d\vec{l}\times\vec{B} = I\,\vec{L}\times\vec{B} \), where \( \vec{L} \) is the vector from the start to the end of the wire.
Reading the figure, the wire goes from A to E: a straight part AB of length \( R \), a semicircle BD of radius \( R \) (so BD is a diameter, \( 2R \)), and a straight part DE of length \( R \).
\( \displaystyle L = AE = R + 2R + R = 4R \)
The field is perpendicular to the plane of the wire, so \( \sin 90^\circ = 1 \).
\( \displaystyle F = B\,I\,L = 4BIR \)