1.

The ratio of magnetic field and magnetic moment at the centre of a current carrying circular loop is x. Whenboth the current and radius is doubled the ratio will be

Answer»

`x/8`
`x/4`
`x/2`
2x

Solution :MAGNETIC field at the centre of a CIRCULAR loop of radius R carrying current I is
`B = (mu_(0)2piI)/(2piR)=(mu_(0)I)/(2R)`
Its magnetic moment is `M = IA = I (piR^(2))`
`therefore B/M = (mu_(0)I)/(2R)xx1/(IpiR^(2))=(mu_(0))/(2piR^(3))=x` [Given]
When both the current and radius is DOUBLED, the ratio becomes
`(B.)/(M.)=(mu_(0))/(2pi(2R)^(3))=1/8(mu_(0)/(2piR^(3)))=x/8`


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