Question

A rectangular wire-loop of width a is suspended from the insulated pan of a spring balance, as shown in the figure. A current i exists in the anti-clockwise direction in the loop. A magnetic field B exists in the lower region. Find the change in the tension of the spring if the current in the loop is reversed.

Answer 1

Given,
A rectangular wire loop of width a
Electric current through the loop = i
Direction of the current is anti-clockwise.
Strength of the magnetic field in the lower region = B
Direction of the magnetic field is into the plane of the loop.



Here, angle between the length of the loop and magnetic field, θ = 90˚
Magnetic force is given by
$\overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B}$
The magnetic force will act only on side AD and BC.
As side AD is outside the magnetic field, so F = 0
Magnetic force on side BC is
$$\begin{gathered} \overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B} \\ =iaB\sin \theta \\ =iaB \end{gathered}$$
Direction of force can be found using Fleming's left-hand rule.
Thus, the direction of the magnetic force is upward.
Similarly if we change the direction of current to clockwise,
the force along BC,
$$\begin{gathered} \overrightarrow{F}=i\overrightarrow{a}\times \overrightarrow{B} \\ =-iaB \end{gathered}$$
Thus, the change in force is equal to the change in tension
= iaB − (− iaB) = 2iaB.