Question

A rigid wire consists of a semi-circular portion of radius R and two straight sections (figure). The wire is partially immersed in a perpendicular magnetic field B, as shown in the figure. Find the magnetic force on the wire if it carries a current i.

Answer 1

Given:
Radius of the semi-circular portion of the rigid wire = R
Magnetic field = B
Electric current flowing through the wire = i
As per the question, the wire is partially immersed in a perpendicular magnetic field.

As PQ and RS are straight wires of length l each and strength of the magnetic field is also same on both the wires, the force acting on these wires will be equal in magnitude but their directions will be opposite to each other.(Direction of force can be found out using Fleming's left hand rule.)
So, the magnetic force on the wire PQ and the force on the wire RS are equal and opposite to each other. Both the forces cancel each other.
Therefore, only the semicircular loop PR will experience a net magnetic force.
Here, angle between the length of the wire and magnetic field, θ = 90˚
Magnetic force in the loop PR,
$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$
Here, l = 2R
$$\begin{gathered} \overrightarrow{F}\mathit{=}\mathit{}i2R\mathit{\times }\overrightarrow{B} \\ \overrightarrow{F}\mathit{=}\mathit{}i2RB\sin 90°=i2RB \end{gathered}$$