Current $I$ flowing along edges of one face of a cube as shown in the figure – I produces magnetic field $\vec{B} = B_{0} \hat{j}$ at the centre of the cube. Consider another identical cube, where the current I flows along the path shown in the figure – II. The magnetic field at the centre of the second cube is $B_{0} [a \overset{\land}{i} + b \overset{\land}{j} + c \overset{\land}{k}]$ . Find $a + b + c$ .

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Detailed Solution

Assume equal and opposite current along sides AE, EF and EH to get 3 square loops.

${\vec{B}}_{c e n t r e} = {\vec{B}}_{A B F E} + {\vec{B}}_{A E H D} + {\vec{B}}_{E F G H} = B_{0} (- \overset{\land}{i}) + B_{0} (\overset{\land}{j}) + B_{0} (\overset{\land}{k})$

Therefore, a + b + c = 1.

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