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Question

Physics Question on Moving charges and magnetism

Three long straight wires A,BA, B and CC are carrying currents as shown in figure. Then the resultant force on BB is directed .............

A

towards A

B

towards C

C

perpendicular to the plane of paper and outward

D

perpendicular to the plane of paper and inward

Answer

towards C

Explanation

Solution

B has attractive force due to A and C both \therefore FAB=μ04π2×1×2d=4π04πdF_{AB} = \frac{\mu_0}{4 \pi } \frac{2 \times 1 \times 2}{d} = \frac{4 \pi_0}{4 \pi d} and FCB=μ04π2×2×3d=12μ04πdF_{CB} = \frac{\mu_0}{4 \pi } \frac{2 \times 2 \times 3}{d} = \frac{12 \,\mu_0}{4 \pi d} Clearly FCBF_{CB} is more effective.