A square of side 2.0m is placed in a uniform magnetic field $B = 2 .0 T$ in a direction perpendicular to the plane of the square inwards. Equal current $i = 3 .0 A$ is flowing in the directions shown in figure. Find the magnitude of the magnetic force on the loop.

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Solution

Since the force on wire ACD, the force on wire AD and the force on wire AED are equal.

The net force on the loop is given as,

$F_{n e t} = 3 {\to F}_{A D}$

$= 3 i \times (A D) \times B$

$= 3 \times 3 \times (2 \sqrt{2}) \times 2$

$= 50.91 N$

Thus, the net force on the loop is $50.91 N$ .

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A square of side 2.0m is placed in a uniform magnetic field B=2.0T in a direction perpendicular to the plane of the square inwards. Equal current i=3.0A is flowing in the directions shown in figure. Find the magnitude of the magnetic force on the loop.

Since the force on wire ACD, the force on wire AD and the force on wire AED are equal. The net force on the loop is given as, Fnet=3→FAD =3i×(AD)×B =3×3×(2√2)×2 =50.91N Thus, the net force on the loop is 50.91N.

A square of side 2.0m is placed in a uniform magnetic field $B = 2 .0 T$ in a direction perpendicular to the plane of the square inwards. Equal current $i = 3 .0 A$ is flowing in the directions shown in figure. Find the magnitude of the magnetic force on the loop.

Since the force on wire ACD, the force on wire AD and the force on wire AED are equal.

The net force on the loop is given as,

$F_{n e t} = 3 {\to F}_{A D}$

$= 3 i \times (A D) \times B$

$= 3 \times 3 \times (2 \sqrt{2}) \times 2$

$= 50.91 N$

Thus, the net force on the loop is $50.91 N$ .