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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 |
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Answer» `x/8` `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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