The frame shown is made of identical rods, each of length $L = 1 m$ . A current of $i = 2 A$ is observed to be flowing through wire $C A$ . The current enters at $C$ and leaves at $D$ . It is placed in a uniform magnetic field $B = 1 T$ , acting perpendicular to the plane of the frame as shown. Find the total magnetic force acting on the frame (in newtons).

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Solution

Given, the identical rods are of equal length.
$R_{C A} = R_{A D} = R_{C E} = R_{E D}$
& $R_{C D} = \frac{1}{2} R_{C A}$
[since $R \propto l$ for identical wire]

Hence, the current flowing in different branchs of the frame will be as follows:

Make a loop $C A D C$ carrying current $2 A$
and a loop $C E D C$ carrying current $2 A$
Then, the rod $C D$ carries current $8 A$

We know that a current carrying loop in a uniform magnetic field experiences no net force.

Hence, force is only on the rod $C D$ carrying $8 A$ .

Total magnetic force, $F = B i L sin θ = 1 \times 8 \times 1 \times sin 90^{\circ} = 8 N$

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