The ratio of the magnetic field at the centre | vedclass

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(A) The magnetic field at the centre of a circular ring carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2R}$.The magnetic field at a point on t

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$27:1$

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(A) The magnetic field at the centre of a circular ring carrying current $I$ is given by $B_c = \frac{\mu_0 I}{2R}$.The magnetic field at a point on the axis of the ring at a distance $x$ from the centre is given by $B_a = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}$.Given $x = 2\sqrt{2}R$, we calculate the term $(R^2+x^2)$:$R^2 + x^2 = R^2 + (2\sqrt{2}R)^2 = R^2 + 8R^2 = 9R^2$.Substituting this into the expression for $B_a$:$B_a = \frac{\mu_0 I R^2}{2(9R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(27R^3)} = \frac{\mu_0 I}{54R}$.Now, the ratio $B_c/B_a$ is:$\frac{B_c}{B_a} = \frac{\mu_0 I}{2R} \div \frac{\mu_0 I}{54R} = \frac{54}{2} = 27$.Therefore, the ratio is $27:1$.

The ratio of the magnetic field at the centre of a circular ring of radius $R$ to the magnetic field at a point on its axis at a distance $2\sqrt{2}R$ from its centre is . . . . . .

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