Question

A current of 2 A enters at the corner d of a square frame abcd of side 20 cm and leaves at the opposite corner b. A magnetic field B = 0.1 T exists in the space in a direction perpendicular to the plane of the frame, as shown in the figure. Find the magnitude and direction of the magnetic forces on the four sides of the frame.

Answer 1

Given:
A square frame abcd of side, l = 20 cm
Electric current through the wire, I = 2 A
Magnetic field, B = 0.1 T
The direction of magnetic field is perpendicular to the plane of the frame, coming out of the plane.
As per the question, current enters at the corner d of the square frame and leaves at the opposite corner b.
Angle between the frame and magnetic field, θ = 90˚
Magnetic force,
$\overrightarrow{F}=i\overrightarrow{l}\times \overrightarrow{B}$
For wire da and cb,
$\left|\overrightarrow{F}\right|$ = ilBsinθ
∴$\left|\overrightarrow{\mathit{F}}\right|$ = ilBsin90˚
= $\frac{2}{2}$ × 20 × 10⁻² × 0.1
= 0.02 N
The direction of force can be found using Fleming's left-hand rule.
Thus, the direction of magnetic force is towards the left.
For wires dc and ab,
$\left|\overrightarrow{F}\right|$ = ilBsinθ
∴$\left|\overrightarrow{\mathit{F}}\right|$ = ilBsin90˚
= $\frac{2}{2}$ × 20 × 10⁻² × 0.1
= 0.02 N
The direction of force can be found using Fleming's left-hand rule.
Thus, the direction of magnetic force is downwards.