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In figure, the cube is 40.0 cm on each edge. Four straight segments of wire $a b, b c, c d$ and $a d$ 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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$F_{a b} = 0, F_{b c} = (- 0.04 N) \hat{i}, F_{c d} = (- 0.04 N) \hat{k}, F_{d a} = (0.04 \hat{i} + 0.04 \hat{k}) N$

$F_{a b} = 0, F_{b c} = (0.04 N) \hat{i}, F_{c d} = (- 0.04 N) \hat{k}, F_{d a} = (0.04 \hat{i} + 0.04 \hat{k}) N$

$F_{a b} = 0, F_{b c} = (- 0.04 N) \hat{i}, F_{c d} = (- 0.04 N) \hat{k}, F_{d a} = (- 0.04 \hat{i} + 0.04 \hat{k}) N$

$F_{a b} = 0, F_{b c} = (0.04 N) \hat{i}, F_{c d} = (0.04 N) \hat{k}, F_{d a} = (0.04 \hat{i} + 0.04 \hat{k}) N$

answer is A.

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Detailed Solution

$i = 5 A \Rightarrow B = (0.02 \hat{j}) T$

Now applying $F = i (I \times B)$ in all parts. Let us find I for anyone parts.

$I_{c d} = r_{d} - r_{c}$

$= (0.4 \hat{j} + 0.4 \hat{k}) - (0.4 \hat{i} + 0.4 \hat{k})$

$= (0.4 \hat{j} - 0.4 \hat{i})$

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