The ratio of the magnetic field at the centre | vedclass

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(A) The magnetic field at the center of a circular coil with $N$ turns,current $I$,and radius $R$ is given by $B = \frac{\mu_0 NI}{2R}$.The magnetic m

Vedclass · Accepted Answer

$\frac{\alpha}{8}$

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(A) The magnetic field at the center of a circular coil with $N$ turns,current $I$,and radius $R$ is given by $B = \frac{\mu_0 NI}{2R}$.The magnetic moment of the coil is $M = NIA = NI(\pi R^2)$.The ratio $\alpha$ is given by $\frac{B}{M} = \frac{\mu_0 NI}{2R} 	imes \frac{1}{NI\pi R^2} = \frac{\mu_0}{2\pi R^3}$.From this expression,we see that $\alpha \propto \frac{1}{R^3}$,and it is independent of the current $I$.If the radius $R$ is doubled $(R' = 2R)$,the new ratio $\alpha'$ becomes:$\alpha' = \frac{\mu_0}{2\pi (2R)^3} = \frac{\mu_0}{2\pi (8R^3)} = \frac{1}{8} \left( \frac{\mu_0}{2\pi R^3} ight) = \frac{\alpha}{8}$.Thus,the new ratio is $\frac{\alpha}{8}$.

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

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