4.6 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR CURRENT LOOP

In this section, we shall evaluate the magnetic field due to a circular coil along its axis. The evaluation entails summing up the effect of infinitesimal current elements ($Id\mathbf{l}$) mentioned in the previous section. We assume that the current $I$ is steady and that the evaluation is carried out in free space (i.e., vacuum).

4.11

Magnetic field on the axis of a current carrying circular loop of radius $R$. Shown are the magnetic field $d\mathbf{B}$ (due to a line element $d\mathbf{l}$ ) and its components along and perpendicular to the axis.

Figure 4.11 depicts a circular loop carrying a steady current $I$. The loop is placed in the $y-z$ plane with its centre at the origin $O$ and has a radius $R$. The $x$-axis is the axis of the loop. We wish to calculate the magnetic field at the point $P$ on this axis. Let $x$ be the distance of $P$ from the centre $O$ of the loop.

Consider a conducting element $d\mathbf{l}$ of the loop. This is shown in Fig. 4.11. The magnitude $dB$ of the magnetic field due to $d\mathbf{l}$ is given by the Biot-Savart law [Eq. 4.11(a)],

$$dB=\frac{{\mu }_0}{4\pi }\frac{I|d\mathit{l}\times \mathbf{r}|}{r^3}$$(4.12)

Now $r_2=x_2+R_2$ . Further, any element of the loop will be perpendicular to the displacement vector from the element to the axial point. For example, the element $d\mathbf{l}$ in Fig. 4.11 is in the $y-z$ plane, whereas, the displacement vector $\mathbf{r}$ from $d\mathbf{l}$ to the axial point $P$ is in the $x-y$ plane. Hence $|d\mathbf{l}\mathbf{\times }\mathbf{r}|=rdl$. Thus,

$$\mathrm{d}B=\frac{{\mu }_0}{4\pi }\frac{I\mathrm{d}l}{(x^2+R^2)}$$(4.13)

The direction of $d\mathbf{B}$ is shown in Fig. 4.11. It is perpendicular to the plane formed by $d\mathbf{l}$ and $\mathbf{r}$. It has an x-component $d{\mathbf{B}}_x$ and a component perpendicular to x-axis, $d{\mathbf{B}}_{\perp }$. When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result. For example, the $d{\mathbf{B}}_{\perp }$ component due to dl is cancelled by the contribution due to the diametrically opposite $d\mathbf{l}$ element, shown in Fig. 4.11. Thus, only the x-component survives. The net contribution along x-direction can be obtained by integrating $d{B}_x=dBcos\theta$ over the loop. For Fig. 4.11,

$$\cos \theta =\frac{R}{{(x^2+R^2)}^{1/2}}$$(4.14)

From Eqs. (4.13) and (4.14), $$\mathrm{d}{B}_x=\frac{{\mu }_{0}I\mathrm{d}l}{4\pi }\frac{R}{{(x^2+R^2)}^{3/2}}$$

The summation of elements $dl$ over the loop yields $2\pi R$, the circumference of the loop. Thus, the magnetic field at $P$ due to entire circular loop is

$$\mathbf{B}=B_x\hat{\mathbf{i}}=\frac{{\mu }_{0}I{R}^2}{2{(x^2+R^2)}^{3/2}}\hat{\mathbf{i}}$$(4.15)

As a special case of the above result, we may obtain the field at the centre of the loop. Here $x=0$, and we obtain,

$${\mathbf{B}}_0=\frac{{\mu }_{0}I}{2R}\hat{\mathbf{i}}$$(4.16)

4.12

The magnetic field lines for a current loop. The direction of the field is given by the right-hand thumb rule described in the text. The upper side of the loop may be thought of as the north pole and the lower side as the south pole of a magnet.

The magnetic field lines due to a circular wire form closed loops and are shown in Fig. 4.12. The direction of the magnetic field is given by (another) right-hand thumb rule stated below:

Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field.

Example 4.6

A straight wire carrying a current of $12A$ is bent into a semi-circular arc of radius $2.0cm$ as shown in Fig. 4.13(a). Consider the magnetic field $\mathbf{B}$ at the centre of the arc.
(a) What is the magnetic field due to the straight segments?
(b) In what way the contribution to $\mathbf{B}$ from the semicircle differs from that of a circular loop and in what way does it resemble?
(c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. 4.13(b)?

4.13

VIEW SOLUTION

1. $d\mathbf{l}$ and $\mathbf{r}$ for each element of the straight segments are parallel. Therefore, $d\mathbf{l}\mathbf{\times }\mathbf{r}=0$. Straight segments do not contribute to |$\mathbf{B}$|.
2. For all segments of the semicircular arc, $d\mathbf{l}\mathbf{\times }\mathbf{r}$ are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of $\mathbf{B}$ for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus $\mathbf{B}$ is $1.9\times {10}^{–4}T$ normal to the plane of the paper going into it.
3. Same magnitude of $\mathbf{B}$ but opposite in direction to that in (b).