A circular coil of radius 10 ,cm with 100 tur | vedclass

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Question

(B) Given: Radius of coil $R = 10 \,cm = 0.1 \,m$,number of turns $N = 100$,current $I = 0.5 \,A$,magnetic field $B = 2 \,T$.The magnetic moment $M$ o

Vedclass · Accepted Answer

$2 \pi \,J$

Vedclass · Answer

(B) Given: Radius of coil $R = 10 \,cm = 0.1 \,m$,number of turns $N = 100$,current $I = 0.5 \,A$,magnetic field $B = 2 \,T$.The magnetic moment $M$ of the coil is given by $M = N I A$,where $A = \pi R^2$.$A = \pi (0.1)^2 = 0.01 \pi \,m^2$.$M = 100 	imes 0.5 	imes 0.01 \pi = 0.5 \pi \,A \cdot m^2$.The work done $W$ in rotating a magnetic dipole in a magnetic field from angle $	heta_1$ to $	heta_2$ is given by $W = MB(\cos 	heta_1 - \cos 	heta_2)$.Here,$	heta_1 = 0^{\circ}$ and $	heta_2 = 180^{\circ}$.$W = (0.5 \pi) 	imes 2 	imes (\cos 0^{\circ} - \cos 180^{\circ})$.$W = \pi 	imes (1 - (-1)) = \pi 	imes 2 = 2 \pi \,J$.Therefore,the correct option is $B$.

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