Question

A current flows through a circular loop of radius r is shown in the figure.

1. What is the direction of magnetic field at ‘o’?

2. Derive an equation for magnetic field at ‘o’ due to the circular loop carrying current i?

3. If the loop splits into two equal halves as shown in figure.

What will be the magnetic field at the center ‘o’?

Answer 1

1. B = \(\frac{\mu_0I}{2r}\) in to the plane of paper 

2. Magnetic field on the axis of a circular current loop:

Consider a circular loop of radius ‘a’ and carrying current T. Let P be a point on the axis of the coil, at distance x from A and r from ‘O’. Consider a small length dl at A. The magnetic field at ‘p’ due to this small element dl,

dB = \(\frac{\mu_0IdI\,sin\,90}{4\pi x^2}\)

dB = \(\frac{\mu_0IdI}{4\pi \,x^2}\)  .............(1)

[since sin 90° -1]

The dB can be resolved into dB cosΦ (along Py) and dB sinΦ (along Px). Similarly consider a small element at B, which produces a magnetic field ‘dB’ at P. If we resolve this magnetic field we get. dB sinΦ (along px) and dB cosΦ (along py ) dB cosΦ components cancel each other, because they are in opposite direction. So only dB sinΦ components are found at P, so total led at P is

B = \(\int dB\,sin\phi\)

= \(\int \frac{\mu_0\,IdI}{4 \pi\,x^2}sin\phi\)

but from ΔAOP we get, sinΦ = a/x

∴ We get

B = \(\int\frac{\mu_0\,Ia}{4\pi\,x^3}dI\)

B = \(\int\frac{\mu_0\,Ia}{4\pi\,x^3}\int dI\)

B = \(\int\frac{\mu_0\,Ia}{4\pi\,x^3}2\pi a\)    [since \(\int dI=2\pi a\)]

B = \(\int\frac{\mu_0\,Ia^2}{2x^3}\)

from ∆AOP we get x = (r² + a²)^1/2

i.e. B = \(\frac{\mu_0\,Ia^2}{2(r^2+a^2)^{3/2}}\)

Let there be N turns in the loop then,

B = \(\frac{\mu_0NIa^2}{2(r^2+a^2)^3/2}\)

Point at the centre of the loop: When the point is at the centre of the loop, (r = 0) Then,

B = \(\frac{\mu_0NI}{2a}\)

B = \(\frac{\mu_0R^2I}{2(x^2+R^2)^{3/2}}\)

3. Zero