5.2 The Bar Magnet

One of the earliest childhood memories of the famous physicist Albert Einstein was that of a magnet gifted to him by a relative. Einstein was fascinated, and played endlessly with it. He wondered how the magnet could affect objects such as nails or pins placed away from it and not in any way connected to it by a spring or string.

We begin our study by examining iron filings sprinkled on a sheet of glass placed over a short bar magnet. The arrangement of iron filings is shown in Fig. 5.2.

The pattern of iron filings suggests that the magnet has two poles similar to the positive and negative charge of an electric dipole. As mentioned in the introductory section, one pole is designated the North pole and the other, the South pole. When suspended freely, these poles point approximately towards the geographic north and south poles, respectively. A similar pattern of iron filings is observed around a current carrying solenoid.

5.2.1 The magnetic field lines

The pattern of iron filings permits us to plot the magnetic field lines*. This is shown both for the bar-magnet and the current-carrying solenoid in Fig. 5.3. For comparison refer to the Chapter 1, Figure 1.17(d). Electric field lines of an electric dipole are also displayed in Fig. 5.3(c). The magnetic field lines are a visual and intuitive realisation of the magnetic field. Their properties are:

1. The magnetic field lines of a magnet (or a solenoid) form continuous closed loops. This is unlike the electric dipole where these field lines begin from a positive charge and end on the negative charge or escape to infinity.
2. The tangent to the field line at a given point represents the direction of the net magnetic field B at that point.
3. The larger the number of field lines crossing per unit area, the stronger is the magnitude of the magnetic field B. In Fig. 5.3(a), B is larger around region ii than in region i .
4. The magnetic field lines do not intersect, for if they did, the direction of the magnetic field would not be unique at the point of intersection. One can plot the magnetic field lines in a variety of ways. One way is to place a small magnetic compass needle at various positions and note its orientation. This gives us an idea of the magnetic field direction at various points in space.

5.2.2 Bar magnet as an equivalent solenoid

In the previous chapter, we have explained how a current loop acts as a magnetic dipole (Section 4.10). We mentioned Ampere’s hypothesis that all magnetic phenomena can be explained in terms of circulating currents. Recall that the magnetic dipole moment m associated with a current loop was defined to be m = NIA where N is the number of turns in the loop, I the current and A the area vector (Eq. 4.30).

To make this analogy more firm we calculate the axial field of a finite solenoid depicted in Fig. 5.4 (a). We shall demonstrate that at large distances this axial field resembles that of a bar magnet.

Let the solenoid of Fig. 5.4(a) consists of n turns per unit length. Let its length be 2l and radius a. We can evaluate the axial field at a point P, at a distance r from the centre O of the solenoid. To do this, consider a circular element of thickness dx of the solenoid at a distance x from its centre. It consists of n dx turns. Let I be the current in the solenoid. In Section 4.6 of the previous chapter we have calculated the magnetic field on the axis of a circular current loop. From Eq. (4.13), the magnitude of the field at point P due to the circular element is

\[ d B=\frac{\mu_{0} n d x I a^{2}}{2\left[(r-x)^{2}+a^{2}\right]^{3 / 2}} \]

The magnitude of the total field is obtained by summing over all the elements — in other words by integrating from x = – l to x = + l. Thus,

\[ B=\frac{\mu_{0} \pi I a^{2}}{2} \int_{-1}^{1} \frac{d x}{\left[(r-x)^{2}+a^{2}\right]^{3 / 2}} \]

This integration can be done by trigonometric substitutions. This exercise, however, is not necessary for our purpose. Note that the range of x is from – l to + l. Consider the far axial field of the solenoid, i.e., r>>a and r>>l. Then the denominator is approximated by

\[ \begin{aligned} \left[(r-x)^{2}+a^{2}\right]^{3 / 2} \approx r^{3} \\ \text { and } B=\frac{\mu_{0} n I a^{2}}{2 r^{3}} \int_{-l}^{1} d x \\ =\frac{\mu_{0} n I}{2} \frac{2 l a^{2}}{r^{3}} \end{aligned} \]

Note that the magnitude of the magnetic moment of the solenoid is, \(m = n (2l) I (\pi a^{2} )\) — (total number of turns × current × cross-sectional area). Thus,

\[ B=\frac{\mu_{0}}{4 \pi} \frac{2 m}{r^{3}} \]

This is also the far axial magnetic field of a bar magnet which one may obtain experimentally. Thus, a bar magnet and a solenoid produce similar magnetic fields. The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.

Some textbooks assign a magnetic charge (also called pole strength) +\(q_{m}\) to the north pole and –\(q_{m}\) to the south pole of a bar magnet of length 2l, and magnetic moment q m (2l). The field strength due to q m at a distance r from it is given by \(\mu_{0} q_{m} / 4 \pi r^{2}\) . The magnetic field due to the bar magnet is then obtained, both for the axial and the equatorial case, in a manner analogous to that of an electric dipole (Chapter 1). The method is simple and appealing. However, magnetic monopoles do not exist, and we have avoided this approach for that reason.

5.2.3 The dipole in a uniform magnetic field

The pattern of iron filings, i.e., the magnetic field lines gives us an approximate idea of the magnetic field B. We may at times be required to determine the magnitude of B accurately. This is done by placing a small compass needle of known magnetic moment m and moment of inertia I I and allowing it to oscillate in the magnetic field. This arrangement is shown in Fig. 5.4(b).

The torque on the needle is [see Eq. (4.29)],

\[ \tau=\mathbf{m} \times \mathbf{B} \]

In magnitude \( \tau=m B \sin \theta \)

Here \( \tau \) is restoring torque and θ is the angle between m and B. Therefore, in equilibrium

\[ \mathscr{J} \frac{d^{2} \theta}{d t^{2}}=-m B \sin \theta \]

Negative sign with mB sin θ implies that restoring torque is in opposition to deflecting torque. For small values of θ in radians, we approximate sin θ ≈ θ and get

\[ \begin{aligned} & y \frac{d^{2} \theta}{d t^{2}} \approx-m B \theta \\ \text { or, } \frac{d^{2} \theta}{d t^{2}} &=-\frac{m B}{y} \theta \end{aligned} \]

This represents a simple harmonic motion. The square of the angular frequency is \(\omega^{2}=m B / \mathscr{J}\) and the time period is,

\[ \begin{array}{l} T=2 \pi \sqrt{\frac{y}{m B}} \\ \text { or } \quad B=\frac{4 \pi^{2} g}{m T^{2}} \end{array} \]

An expression for magnetic potential energy can also be obtained on lines similar to electrostatic potential energy. The magnetic potential energy \(U_{m}\) is given by

\[ \begin{aligned} U_{m} &=\int \tau(\theta) d \theta \\ &=\int m B \sin \theta d \theta=-m B \cos \theta \\ &=-\mathbf{m} \cdot \mathbf{B} \end{aligned} \]

We have emphasised in Chapter 2 that the zero of potential energy can be fixed at one’s convenience. Taking the constant of integration to be zero means fixing the zero of potential energy at θ = 90°, i.e., when the needle is perpendicular to the field. Equation (5.6) shows that potential energy is minimum (= –mB) at θ = 0° (most stable position) and maximum (= +mB) at θ = 180° (most unstable position).

5.2.4 The electrostatic analog

Comparison of Eqs. (5.2), (5.3) and (5.6) with the corresponding equations for electric dipole (Chapter 1), suggests that magnetic field at large distances due to a bar magnet of magnetic moment m can be obtained from the equation for electric field due to an electric dipole of dipole moment p, by making the following replacements:

\[ \mathbf{E} \rightarrow \mathbf{B}, \mathbf{p} \rightarrow \mathbf{m}, \frac{1}{4 \pi \varepsilon_{0}} \rightarrow \frac{\mu_{0}}{4 \pi} \]

In particular, we can write down the equatorial field (\(B_{E}\) ) of a bar magnet at a distance r, for r >> l, where l is the size of the magnet:

\[ \mathbf{B}_{E}=-\frac{\mu_{0} \mathbf{m}}{4 \pi r^{3}} \]

Likewise, the axial field (\(B_{A}\) ) of a bar magnet for r >> l is:

\[ \mathbf{B}_{A}=\frac{\mu_{0}}{4 \pi} \frac{2 \mathbf{m}}{r^{3}} \]

Equation (5.8) is just Eq. (5.2) in the vector form. Table 5.1 summarises the analogy between electric and magnetic dipoles.