A) zero
B) \[0.5\times {{10}^{-4}}A{{m}^{2}}\]
C) \[1.25\times {{10}^{-4}}A{{m}^{2}}\]
D) magnetic moment is not defined for this case
Correct Answer: B
Solution :
A moving charge along a circle is equivalent to a current carrying ring, whose current is given by, \[I=\frac{q}{T}=\frac{q}{2\pi /\omega }\] \[\Rightarrow \] \[I=\frac{4\times {{10}^{-6}}}{2\pi }\times 100\] \[=0.64\times {{10}^{-4}}A\] Magnetic moment of rotating charge is \[M=IA\] \[\Rightarrow \] \[M=0.64\times {{10}^{-4}}\times \pi \times {{0.5}^{2}}\] \[=0.5\times {{10}^{-4}}A-{{m}^{2}}\]You need to login to perform this action.
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