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The dipole moment of a coil of 200 turns carrying a current of `3A` is `9*24Am^2`. What is the diameter of the coil? |
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Answer» Correct Answer - `14cm` Here, `N=200`, `I=3A`, `M=9*24Am^2d=?` If r is radius of coil, then `M=NIA=Nipi(d^2)/(4)` `d=sqrt((4M)/(piNI))=sqrt((4xx9*24)/(3*14xx200xx3))` `=0*14m=14cm` |
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