Question

A wire carrying current I, bent as shown in the figure, is placed in a uniform magnetic 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.

• A. BI(2R + πR), vertically downward
• B. 3BIR, directed vertically upward
• C. 4BIR, directed vertically downward
• D. 2πBIR, from P to Q

Answer 1

The Correct Option is A

The force on a current-carrying wire segment in a uniform magnetic field is calculated using F = IBLsinθ, where I represents current, B is the magnetic field strength, L is the wire length, and θ is the angle between the wire and the field. In this case, the magnetic field is perpendicular to the wire, making sinθ = 1. The wire comprises two straight sections and a semicircular arc.

• Each straight section has a length of 2R, equaling the circle's radius.
• The semicircular arc has a radius R, resulting in a length of πR.

The total wire length L is the sum of these parts:
\(L = 2R + πR\)
Consequently, the total magnetic force on the wire is:
\(F = IB(2R + πR)\)
The force direction, determined by the right-hand rule with the magnetic field directed out of the plane, is vertically downward. The final answer is therefore \(BI(2R + πR)\), vertically downward.