InterviewSolution
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InterviewSolution
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A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. A current I passes through the coil. The magnetic moment of the spriral isA. `M=(mu_(0)NI)/(3)(b^(3)-a^(3))/((b-a)^(3))`B. `M=(piNI)/(3)((b-a)^(3))/(b^(3)-a^(3))`C. `M=(piNI)/(3)((b^(3)-a^(3)))/((b-a)^(3))`D. `M=(mu_(0)NI)/(3)((b-a)^(3))/(b^(3)-a^(3))` |
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Answer» Correct Answer - C Magnetic moment for one loop of radius r is `M=IA=I(pir^(2))` Total number of turns per unit length is `(N)/(b-a)` `dM=(dN)(IA)dM=((N)/(b-a))dr` `(pi r^(2)I)=(piNI)/(b-a)r^(2)dr` `M=(piNI)/(b-a)int_(a)^(b)r^(2)dr=(piNI)/(b-a)(r^(3)/(3))_(a)^(b)=(piNI)/(3)((b^(3)-a^(3))/(b-a))` |
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