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The ratio of the magnetic field at the centre of a current carrying coil to its magnetic moment is $ x$. If the current and the radius are both doubled, the new ratio will become

$\begin {array} {1 1} (a)\;8 \: x & \quad (b)\;4\: x \\ (c)\;\large\frac{x}{8} & \quad (d)\;\large\frac{x}{4} \end {array}$


1 Answer

$x = \large\frac{B}{M}$
On substituting the expressions for $B$ and $M$, one obtains that $x$
is independent of current and is proportional to $r ^ {-3}$
Ans : (c)
answered Feb 19, 2014 by thanvigandhi_1

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