Question

The diagram show cross-section of several conductors that carry currents perpendicular to the plane of the diagram. The currents have magnitudes I1=3.5 A,I2=6.5 A, I3=1.5 A and their directions are as shown. Four closed paths labelled a to d are shown. The line integral ∮→B⋅→dl for the paths a, b, c and d will be respectively (Each integral going around the path in counterclockwise direction).

A
0, 3.5μo, 3μo, 4.5μo
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B
0, 3.5μo, 3μo, 4.5μo
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C
0, 3μo, 3.5μo, 11.5μo
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D
All integrals will be equal to zero.
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Solution

The correct option is A 0, −3.5μo, 3μo, 4.5μoAccording to Ampere's law, ∮→B⋅→dl=μo⟮∑I⟯ From the right-hand thumb rule, when going around the path in a counter-clockwise direction, currents out of the page are positive and currents into the page are negative. For path a, ∑I=0 For path b, ∑I=−I1=−3.5 A For path c, ∑I=−I1+I2=3 A For path d, ∑I=−I1+I2+I3=4.5 A ⇒∮a→B⋅→dl=0 ⇒∮b→B⋅→dl=−3.5μ0 ⇒∮c→B⋅→dl=3μ0 ⇒∮d→B⋅→dl=4.5μ0 Hence, the correct answer is option (a).

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