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Current I flows through the circuit, as shown in figure. Find the magnetic moment of the figure, if AB = BC = CD = DE= EF = FG=GH=HA.

a
72Iπa2
b
52Iπa2
c
4Iπa2
d
53Iπa2

detailed solution

Correct option is B

Angle θ for one segment, θ=2π8=π4 M0=IA=IA=Ir2θ2-r12θ2 =I(2a)2θ2-a2θ2=I·3a2θ2  For 4 segments, M=4I·3a2θ2=6Ia2θ  Magnetic moment due to circle =Iπa2  Total magnetic moment =6Ia2θ+Iπa2 =6Ia2×π4+I·πa2=52Iπa2

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Similar Questions

A non-conducting non-magnetic rod having circular cross section of radius R is suspended from a rigid support as shown in the figure. A light and small coil of 300 turns is wrapped tightly at the left end of the rod where uniform magnetic field B exists in vertically downward direction.Air of density ρ hits the half of the right part of the rod with velocity V as shown in the figure. What should be current in clockwise direction (as seen from O) in the coil so that rod remains horizontal? Give answer in m A. Given

2LvπRBρ=15 A-1/2


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