A wire carrying current I, bent as shown in the figure, is placed in a uniform field B that emerges normally out from the plane of the figure. The force on this wire is ______________.
Fill in the blank with the correct answer from the options given below.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 4BIR \), directed vertically downward
In a uniform magnetic field, the force on a current-carrying wire of any shape equals the force on a straight wire joining its ends, because \( \vec{F} = I\,\vec{L} \times \vec{B} \) depends only on the vector from the start to the end of the wire.
Readings from the figure (the straight parts are R each and the semicircle has diameter 2R):
\( \displaystyle L_{PQ} = R + 2R + R = 4R \)
\( \vec{L} \) points from P to Q (to the right) and \( \vec{B} \) points out of the plane, so the angle between them is \( 90^\circ \):
\( \displaystyle F = I\,L_{PQ}\,B = 4BIR \)
Direction: \( \vec{L} \times \vec{B} \) with \( \vec{L} \) to the right and \( \vec{B} \) out of the page points vertically downward (in the plane of the figure).