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 (\( I d\mathbf{l} \)) 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)],

\[ d B=\frac{\mu_{0}}{4 \pi} \frac{I|d \boldsymbol{l} \times \mathbf{r}|}{r^{3}} \](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 \( |d\mathbf{l × r}|=r dl \). Thus,

\[ \mathrm{d} B=\frac{\mu_{0}}{4 \pi} \frac{I \mathrm{~d} l}{\left(x^{2}+R^{2}\right)} \](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 \( dB_x=dB cos\theta \) over the loop. For Fig. 4.11,

\[ \cos \theta=\frac{R}{\left(x^{2}+R^{2}\right)^{1 / 2}} \](4.14)

From Eqs. (4.13) and (4.14), \[ \mathrm{d} B_{x}=\frac{\mu_{0} I \mathrm{~d} l}{4 \pi} \frac{R}{\left(x^{2}+R^{2}\right)^{3 / 2}} \]

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

\[ \mathbf{B}=B_{x} \hat{\mathbf{i}}=\frac{\mu_{0} I R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} \hat{\mathbf{i}} \](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,

\[ \mathbf{B}_{0}=\frac{\mu_{0} I}{2 R} \hat{\mathbf{i}} \](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 \( 12 A \) is bent into a semi-circular arc of radius \( 2.0 cm \) 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, \( d\mathbf{l × r} = 0 \). Straight segments do not contribute to |\( \mathbf{B} \)|.
2. For all segments of the semicircular arc, \( d\mathbf{l × r} \) 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 \( 1.9 × 10^{–4} T \) normal to the plane of the paper going into it.
3. Same magnitude of \( \mathbf{B} \) but opposite in direction to that in (b).