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Ratio of the magnetic field at the centre of a current carrying coil of radius $R$ and at a distance of $3R$ on its axis is

$\begin {array} {1 1} (a)\;10\sqrt{10} & \quad (b)\;20\sqrt{10} \\ (c)\;2\sqrt{10} & \quad (d)\;\sqrt{10} \end {array}$

 

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1 Answer

The desired ratio is $ \Large\frac{\Large\frac{\mu_0i}{2R}}{\Large\frac{\mu_0iR^2}{2(R^2+9R^2)^{\Large\frac{3}{2}}}}$
Ans : (a)
answered Feb 19, 2014 by thanvigandhi_1
 

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