(a) Calculate the magnetic field produced at a point along the axis of the current carrying circular coil. Wri

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Q. (a) Calculate the magnetic field produced at a point along the axis of the current carrying circular coil. Write down the equation of the magnetic field at the center of the coil using Biot-Savart law. OR

(b) Derive the equation for angle of deviation produced by a prism and thus obtain the equation for refractive index of material of the prism.

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2024Subjective· 5mImportance★★★★★

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Concept understanding — Magnetic Field on the Axis of a Loop

Magnetic Field on the Axis of a Current Loop

Imagine a circular wire carrying a steady current. You want to know the magnetic field not at the centre, but at some point along the line that passes through the centre and is perpendicular to the plane of the loop — that's the axis.

Why would the field be along the axis at all? Because of symmetry. For every tiny segment of the loop, there is an opposite segment on the other side. Their perpendicular components of the magnetic field cancel out, leaving only the component along the axis. So the net field points straight along the axis, either towards or away from the loop depending on the current direction.

The Intuition

At the centre of the loop ( $x = 0$ ), every segment is at the same distance $R$ from the centre, and the field is strongest. As you move away along the axis, two things happen: the distance from each current element to your observation point increases, and the angle at which the field points along the axis becomes less favourable. So the field drops off.

Far away from the loop, the loop looks like a tiny magnetic dipole — a small bar magnet. The field falls off as $1/x^3$ , exactly like a dipole field.

The Precise Statement

For a circular loop of radius $R$ , carrying a steady current $I$ , the magnitude of the magnetic field at a point on the axis at a distance $x$ from the centre is:

B = \frac{\mu_0 I R^2}{2\,(R^2 + x^2)^{3/2}}

where $\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}$ is the permeability of free space.

The direction of $\mathbf{B}$ is along the axis, given by the right-hand rule: curl the fingers of your right hand in the direction of the current, and your thumb points in the direction of the magnetic field on the axis.

Special Cases

At the centre ( $x = 0$ ):

B_{\text{centre}} = \frac{\mu_0 I}{2R}

Far away ( $x \gg R$ ):

The denominator $(R^2 + x^2)^{3/2} \approx x^3$ , so

B \approx \frac{\mu_0 I R^2}{2 x^3}

This is exactly the field of a magnetic dipole of moment $m = I \cdot (\pi R^2) = I A$ , where $A$ is the area of the loop. So a current loop behaves like a magnetic dipole at large distances.

Watch out

The formula $B = \mu_0 I R^2 / 2(R^2 + x^2)^{3/2}$ is valid only on the axis. Off-axis, the field is much more complicated and cannot be written in such a simple closed form.

Why the $3/2$ Power? …

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(a) The Biot-Savart law, integrated around a circular current loop, gives the on-axis field $B=\mu_0IR^2/[2(R^2+x^2)^{3/2}]$ , reducing to $\mu_0I/2R$ at the centre; (b) a prism's minimum-deviation condition gives its refractive index as $n=\sin[(A+D_m)/2]/\sin(A/2)$ . Both alternatives answered below.

(a) Magnetic field on the axis of a circular coil

1. Setup. Consider a circular coil of radius $R$ carrying current $I$ . We find the field $B$ at a point P on its axis, at distance $x$ from the centre O.

2. Biot-Savart law. For a current element $Id\vec l$ on the coil, the field it produces at P has magnitude

$dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin90°}{r^2} = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl}{(R^2+x^2)}$

since $d\vec l \perp \vec r$ for every element (the current element is tangential, $\vec r$ from element to P), and $r=\sqrt{R^2+x^2}$ .

3. Resolving components. $d\vec B$ is perpendicular to $\vec r$ , and can be resolved into a component along the axis ( $dB\cos\theta$ ) and one perpendicular to it ( $dB\sin\theta$ ), where $\cos\theta=R/\sqrt{R^2+x^2}$ . By symmetry, the perpendicular components from diametrically opposite elements cancel, leaving only the axial components, which add.

4. Integration.

$B = \oint dB\cos\theta = \dfrac{\mu_0 I}{4\pi(R^2+x^2)}\times\dfrac{R}{\sqrt{R^2+x^2}}\oint dl = \dfrac{\mu_0IR}{4\pi(R^2+x^2)^{3/2}}\times2\pi R$

$B = \dfrac{\mu_0IR^2}{2(R^2+x^2)^{3/2}}$

5. At the centre ( $x=0$ ).

$B = \dfrac{\mu_0IR^2}{2R^3} = \dfrac{\mu_0I}{2R}$

(b) Angle of deviation and refractive index of a prism

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Q.(a) Calculate the magnetic field produced at a point along the axis of the current carrying circular coil. Write down the equation of the magnetic field at the center of the coil using Biot-Savart law. OR

(a) Magnetic field on the axis of a circular coil

(b) Angle of deviation and refractive index of a prism

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