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

To solve the problem of deriving the force on the wire in a uniform magnetic field, we use the formula for magnetic force on a segment of a current-carrying wire which is given by the equation: F = IBLsinθ, where I is the current, B is the magnetic field, L is the length of the wire, and θ is the angle between the wire and the magnetic field. In this scenario, the magnetic field is perpendicular to the wire, so sinθ = 1. The wire is bent in an arc of circle combined with two straight sections.

• The length of the straight sections is 2R since both sections are equal to the radius of the circle.
• The arc part is a semicircle with a radius R, thus its length is πR.

Therefore, the total length L of the wire is: 
\(L = 2R + πR\)
Thus, the total force on the wire is:
\(F = IB(2R + πR)\)
The direction of the force is determined using the right-hand rule, considering the convention that the magnetic field emerges out from the plane, leading to a force directed vertically downward. Hence, the solution to the problem is \(BI(2R + πR)\), vertically downward.