In figure,the cube is 40.0 cm on each edge.Four straight segments of wire ab,bc,cd and da form a closed loop that carries a current I=5.00 A,in the direction shown.A uniform magnetic field of magnitude B = 0.020 T is in the positive y-direction.Determine the magnitude and direction of the magnetic force on each segment.

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Solution

We know that,
Length of side = $40 c m = 0.4 m$
Current, $I = 5.00 A$
$B = 0.020 T$
Using, $F = B I L sin θ$
Side l(m) $θ$ F direction
ab 0.4 $π$ 0 -
bc 0.4 $π / 2$ 0.04 (-ve) x axis
cd 0.4 $\sqrt{2}$ $π / 4$ 0.04 (-ve) z axis
da 0.4 $\sqrt{2}$ $π / 4$ 0.566 1/x-z plane

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Side	l(m)	$θ$	F	direction
ab	0.4	$π$	0	-
bc	0.4	$π / 2$	0.04	(-ve) x axis
cd	0.4 $\sqrt{2}$	$π / 4$	0.04	(-ve) z axis
da	0.4 $\sqrt{2}$	$π / 4$	0.566	1/x-z plane

We know that,Length of side = 40cm=0.4mCurrent, I=5.00AB=0.020TUsing, F=BILsinθ Sidel(m) θ F direction ab 0.4 π 0 - bc 0.4 π/2 0.04 (-ve) x axis cd 0.4√2 π/4 0.04 (-ve) z axis da 0.4√2 π/4 0.566 1/x-z plane

We know that,
Length of side = $40 c m = 0.4 m$
Current, $I = 5.00 A$
$B = 0.020 T$
Using, $F = B I L sin θ$
Side l(m) $θ$ F direction
ab 0.4 $π$ 0 -
bc 0.4 $π / 2$ 0.04 (-ve) x axis
cd 0.4 $\sqrt{2}$ $π / 4$ 0.04 (-ve) z axis
da 0.4 $\sqrt{2}$ $π / 4$ 0.566 1/x-z plane