Field On The Axis Of A Circular Loop

Question

A circular loop of radius $R$ carries current $I$; the magnetic field at an axial point located a distance $R$ from the centre compares with the field at the centre as

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From NCERT Class 12, Chapter 4 (Moving Charges and Magnetism), the axial field of a circular loop at distance $x$ from its centre is $B_{axial} = \dfrac{\mu_0 I R^{2}}{2(R^{2}+x^{2})^{3/2}}$, while the central field ($x=0$) is $B_{centre} = \dfrac{\mu_0 I}{2R}$. Taking the ratio, $\dfrac{B_{axial}}{B_{centre}} = \dfrac{R^{3}}{(R^{2}+x^{2})^{3/2}}$. At $x = R$, the denominator becomes $(R^{2}+R^{2})^{3/2} = (2R^{2})^{3/2} = 2\sqrt{2}\,R^{3}$, so the ratio is $\dfrac{R^{3}}{2\sqrt{2}\,R^{3}} = \dfrac{1}{2\sqrt{2}} \approx 0.354$. The '$1/2$' option assumes a simple inverse-square drop, which is wrong. The 'equal' option ignores the distance dependence. The '$1/4$' option misuses the exponent. Plausibility check: the axial field must be smaller than the central field but not negligibly so at a distance equal to the radius, and $\dfrac{1}{2\sqrt{2}} \approx 0.35$ lies sensibly between 0 and 1. As a further test of the formula, at very large distances $x \gg R$ the expression reduces to $B \propto 1/x^{3}$, the characteristic dipole fall-off, which is exactly what a current loop should behave like when viewed from far away, so the general axial formula used here is consistent with the known dipole limit and the intermediate value at $x = R$ is trustworthy.

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Field On The Axis Of A Circular Loop

Select the correct option:

🔒 Solution Hidden from View

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Solution

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In a region with a uniform electric field of $3 \times 1 0^{4}$ V/m perpendicular to a magnetic field...

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