For a circular coil of radius R and N turns carrying current I,
Question:

For a circular coil of radius and turns carrying current I, the magnitude of the magnetic field at a point on its axis at a distance from its centre is given by,

$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{\frac{3}{2}}}$

(a) Show that this reduces to the familiar result for field at the centre of the coil.

(b) Consider two parallel co-axial circular coils of equal radius R, and number of turns N, carrying equal currents in the same direction, and separated by a distance R. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to R, and is given by,

$B=0.72-\frac{\mu_{0} B N I}{R}$, approximately.

[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]

Solution:

Radius of circular coil = R

Number of turns on the coil = N

Current in the coil = I

Magnetic field at a point on its axis at distance x is given by the relation,

$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{\frac{3}{2}}}$

Where,

$\mu_{0}=$ Permeability of free space

(a) If the magnetic field at the centre of the coil is considered, then x = 0.

$\therefore B=\frac{\mu_{0} I R^{2} N}{2 R^{3}}=\frac{\mu_{0} I N}{2 R}$

This is the familiar result for magnetic field at the centre of the coil.

(b) Radii of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both coils = I

Distance between both the coils = R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of $\frac{R}{2}+d$ from point $Q$.

$\therefore$ Magnetic field at point $Q$ is given as:

$B_{1}=\frac{\mu_{0} N I R^{2}}{2\left[\left(\frac{R}{2}+d\right)^{2}+R^{2}\right]^{\frac{3}{2}}}$

Also, the other coil is at a distance of $\frac{R}{2}-d$ from point $Q$.

$\therefore$ Magnetic field due to this coil is given as:

$B_{2}=\frac{\mu_{0} N I R^{2}}{2\left[\left(\frac{R}{2}-d\right)^{2}+R^{2}\right]^{\frac{3}{2}}}$

Total magnetic field,

$B=B_{1}+B_{2}$

$=\frac{\mu_{0} I R^{2}}{2}\left[\left\{\left(\frac{R}{2}-d\right)^{2}+R^{2}\right\}^{-\frac{3}{2}}+\left\{\left(\frac{R}{2}+d\right)^{2}+R^{2}\right\}^{-\frac{3}{2}}\right]$

$=\frac{\mu_{0} I R^{2}}{2}\left[\left(\frac{5 R^{2}}{4}+d^{2}-R d\right)^{-\frac{3}{2}}+\left(\frac{5 R^{2}}{4}+d^{2}+R d\right)^{-\frac{3}{2}}\right]$

$=\frac{\mu_{0} I R^{2}}{2} \times\left(\frac{5 R^{2}}{4}\right)^{-\frac{3}{2}}\left[\left(1+\frac{4}{5} \frac{d^{2}}{R^{2}}-\frac{4}{5} \frac{d}{R}\right)^{-\frac{3}{2}}+\left(1+\frac{4}{5} \frac{d^{2}}{R^{2}}+\frac{4}{5} \frac{d}{R}\right)^{-\frac{3}{2}}\right]$

For $d<<R$, neglecting the factor $\frac{d^{2}}{R^{2}}$, we get:

$\approx \frac{\mu_{0} I R^{2}}{2} \times\left(\frac{5 R^{2}}{4}\right)^{-\frac{3}{2}} \times\left[\left(1-\frac{4 d}{5 R}\right)^{-\frac{3}{2}}+\left(1+\frac{4 d}{5 R}\right)^{-\frac{3}{2}}\right]$

$\approx \frac{\mu_{0} I R^{2} N}{2 R^{3}} \times\left(\frac{4}{5}\right)^{\frac{3}{2}}\left[1-\frac{6 d}{5 R}+1+\frac{6 d}{5 R}\right]$

$B=\left(\frac{4}{5}\right)^{\frac{3}{2}} \frac{\mu_{0} I N}{R}=0.72\left(\frac{\mu_{0} I N}{R}\right)$

Hence, it is proved that the field on the axis around the mid-point between the coils is uniform.

Administrator

Leave a comment

Please enter comment.
Please enter your name.