# Magnetic Field on the Axis of a Circular Coil

Consider a circular loop of radius $r$ carrying a current $I$. The magnetic field at a point P on the axis of the coil at a distance $x$ from its centre is given by \begin{align} B_\mathrm{P}=\frac{\mu_0 I r^2}{2\left(r^2+x^2\right)^{3/2}}. \end{align} Here, $\mu_0$ is the permeability of free space.

The magnetic field is maximum at the coil's centre and it decrease as we go away from the centre. The slope of the magnetic field changes its sign at $x=r/2$. This is an iflection point with $\mathrm{d}^2B/\mathrm{d}x^2=0$.

The magnetic field at the centre of the circular coil is \begin{align} B_\mathrm{C}=\frac{\mu_0 I}{2r}. \end{align}

The magnetic field at the centre of an arc of radius $r$ subtending an angle $\theta$ is \begin{align} B_\mathrm{C}=\frac{\mu_0 I\theta}{4\pi r} \end{align}

## Helmholtz coils

If two similar coils be placed with their axis coincident and separated by a distance $r$, and if same current is passed through them in the same direction, the rate of increase of field due to one coil at midpoint between the coil is equal to the rate of decrease of field due to the other coil. As we move along the axis from the midpoint, decrease in the magnitude of the field due to one coil is exactly compensated by the increase in the field due to the other coil so that the field between the coils is practically uniform. This arrangement of two coils is called Helmholtz coils. Helmholtz coil is used in experiments where a uniform magnetic field is required.

A Helmholtz coil has a pair of loops, each with $N$ turns and radius $r$. They are placed coaxially at distance $r$ and the same current $I$ flows through the loops in the same direction. Find the magnitude of magnetic field at P, midway between the centres A and C (see figure).

The magnetic field at the point P due to both loops are equal. Thus, the resultant magnetic field at P is two times the field by each loop i.e., \begin{align} B&=2\times\frac{N\mu_0 Ir^2}{2(r^2+x^2)^{3/2}} \\ &=\frac{N\mu_0 I r^2}{(r^2+(r/2)^2)^{3/2}}\nonumber\\ &=\frac{8N\mu_0I}{5^{3/2}r}.\nonumber \end{align} The field is almost constant (uniform) in the vicinity of P. Use Helmholtz coil if you need a uniform field.

## Problems from IIT JEE

Problem (IIT JEE 2011): A long insulated copper wire is closely wound as a spiral of $N$ turns. The spiral has inner radius $a$ and outer radius $b$. The spiral lies in $x\text{-}y$ plane and a steady current $I$ flows through the wire. The $z$ component of magnetic field at the centre of spiral is

1. $\frac{\mu_0 N I}{2(b-a)}\ln\big(\frac{b}{a}\big)$
2. $\frac{\mu_0 N I}{2(b-a)}\ln\big(\frac{b+a}{b-a}\big)$
3. $\frac{\mu_0 N I}{2b}\ln\big(\frac{b}{a}\big)$
4. $\frac{\mu_0 N I}{2b}\ln\big(\frac{b+a}{b-a}\big)$

Solution: Consider a small circular element of thickness $\text{d}r$ at a distance $r$ from the centre of spiral (see figure).

The number of turns in this element is $\frac{N}{b-a}\text{d}r$ and thus current flowing through this element is $i=\frac{N}{b-a}\text{d}rI$ (current $I$ through each turn). The magnetic field at the centre of spiral due to this element is given by \begin{align} \mathrm{d}B & =\frac{\mu_0 i}{2r} \\ &=\frac{\mu_0 NI}{2(b-a)}\frac{\text{d}r}{r}. \nonumber \end{align} Integrate $\mathrm{d}B$ from $r=a$ to $r=b$ to get, \begin{align} B&=\int_{a}^{b}\frac{\mu_0 NI}{2(b-a)}\frac{\text{d}r}{r} \\ &=\frac{\mu_0 NI}{2(b-a)}\ln\left(\frac{b}{a}\right). \nonumber \end{align}

## Experimental demo

The magnetic field of a current-carrying circular coil is very interesting. On the coil axis, the field direction is parallel to the axis. On other points in a plane containing coil axis and perpendicular to the coil plane, the magnetic lines of forces form curved lines. See it yourself.

Take enameled copper wire of AWG 18 and wrap about 20-25 turns on a cylindrical object like glass or bottle. The diameter could be about 5 cm. Remove the coil from the cylindrical object and tie it at 3-4 places with the help of cellophane tape or thread. Leave about 10 cm of the wire at each free end of the coil. Fix this coil in vertical position to a flat base such as plastic sheet such that half the coil is inside and other half is outside (see figure). You can use M-seal or any other adhesive to do so. Remove enamel from two free ends of the coil by scratching it with a knife. Paste a white paper on the flat base.

Connect this coil to the DC power supply. Take a magnetic needle and draw the magnetic lines of forces.

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