1.

A rod of length l having uniformly distributed charge Q is rotated about one end with constant frequency 'f'. Its magnetic moment is (pi fQl^(2))/n Then the value of n is

Answer»

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Solution :Charge on the differential element dx,
`dq=Q/L.dx`
Equivalent CURENT di = f dq
`therefore` Magnetic moment of this element
`DMU = (di)NA (N = 1)`
`dpi = (pir^(2))fQ/l dx`
`implies mu = (pifQ)/l int_(0)^(l) X^(2) dx implies mu = 1/3pifQl^(2)`


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