The ratio of the magnetic field at the centre | vedclass

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(D) The magnetic field $B$ at the centre of a circular coil of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.The magnetic dipol

Vedclass · Accepted Answer

$\frac{x}{8}$

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(D) The magnetic field $B$ at the centre of a circular coil of radius $r$ carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.The magnetic dipole moment $m$ is given by $m = I A = I (\pi r^2)$.The initial ratio is $x = \frac{B}{m} = \frac{\mu_0 I / 2r}{I \pi r^2} = \frac{\mu_0}{2 \pi r^3}$.When the current $I$ becomes $I' = 2I$ and the radius $r$ becomes $r' = 2r$,the new magnetic field $B'$ is $B' = \frac{\mu_0 (2I)}{2(2r)} = \frac{\mu_0 I}{2r} = B$.The new magnetic moment $m'$ is $m' = (2I) \pi (2r)^2 = (2I) \pi (4r^2) = 8 I \pi r^2 = 8m$.The new ratio $x'$ is $x' = \frac{B'}{m'} = \frac{B}{8m} = \frac{1}{8} \left( \frac{B}{m} \right) = \frac{x}{8}$.

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

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