Question

A circular loop of radius a, carrying a current i, is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field (figure). The strength of the magnetic field at the periphery of the loop is B. Find the magnetic force on the wire.

Answer 1

Given:
A circular loop of radius = a
So, the length of the loop, l = 2πa
Electric current through the loop = i
As per the question,
The loop is placed in a two-dimensional magnetic field. The centre of the loop coincides with the centre of the field. The strength of the magnetic field at the periphery of the loop is B
Therefore, the magnetic field points radially outwards.
Here, the angle between the length of the loop and the magnetic field, θ = 90˚
Magnetic force is given by
$$\begin{gathered} \overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B} \\ \overrightarrow{F}\mathit{=}\mathit{}i\left(2\mathrm{\pi }a\mathit{\times }\overrightarrow{B}\right) \\ \overrightarrow{F}\mathit{=}\mathit{}i2\mathrm{\pi }aB\sin 90° \\ =i2\mathrm{\pi }aB \end{gathered}$$



Direction of the force can be found using Fleming's left-hand rule.
Thus, the direction of magnetic force is perpendicular to the plane of the figure and pointing inside.