Question

Derive an expression for the magnetic field at a point on the axis of a current carrying circular loop.

Accepted Answer

Let current carrying circular loop of radius $R$ , carrying current $I$ is placed in $y z$ plane and point $P$ is place on axis of loop (i.e. $x$ -axis) at a distance $x$ from centre of loop $O$ as shown in figure.

Consider element $d l$ of loop as shown in figure. Let magnetic field due to this element is $d B$ .

By applying Biot-Savart rule,

$d B = \frac{μ_{0}}{4 π} \frac{I | d l \times r |}{r^{3}} = \frac{μ_{0}}{4 π} \frac{I d l s i n φ}{r^{2}}$

where $ϕ$ is the angle between $d l$ and the displacement vector $r$ .

Since point is on the axis of loop therefore displacement vector $r$ is always perpendicular to element on loop $d l$ .

$∴ ϕ = 90 ° \Rightarrow \sin ϕ = 1$

By Pythagoras theorem, $r^{2} = R^{2} + x^{2}$

$∴ d B = \frac{μ_{0}}{4 π} \frac{I d l}{(x^{2} + R^{2})}$

As we see in figure magnetic field $d B$ has component $d B_{x}$ along $x -$ axis and component $d B_{y}$ along $y -$ axis

$d B_{y} = d B \sin θ$ and $d B_{x} = d B \cos θ$ ,

where $θ$ is angle made by $d B$ vector and with axis of loop ( $x -$ axis)

Net sum of component of magnetic field along $y -$ axis will be zero. Hence, net magnetic field due to all elements of loop at point $P$ will be equal to sum of component of magnetic field along $x -$ axis.

$∴ B = \int_{0}^{2 π R} d B \cos θ = \int_{0}^{2 π R} \frac{μ_{0}}{4 π} \frac{I \cos θ}{(x^{2} + R^{2})} d l = \frac{μ_{0}}{4 π} \frac{I \cos θ}{(x^{2} + R^{2})} \int_{0}^{2 π R} d l$

$\Rightarrow B = \frac{μ_{0} I R \cos θ}{2 (x^{2} + R^{2})} \hat{i}$

Now, $\cos θ = \frac{R}{{(x^{2} + R^{2})}^{1 / 2}}$

$∴ B = \frac{μ_{0} I R^{2}}{2 {(x^{2} + R^{2})}^{3 / 2}} \hat{i}$

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