4.9 Torque on Current Loop, Magnetic Dipole

4.9.1 Torque on a rectangular current loop in a uniform magnetic field

We now show that a rectangular loop carrying a steady current \(I\) and placed in a uniform magnetic field experiences a torque. It does not experience a net force. This behaviour is analogous to that of electric dipole in a uniform electric field (Section 1.12).

We first consider the simple case when the rectangular loop is placed such that the uniform magnetic field \(\mathbf{B}\) is in the plane of the loop. This is illustrated in Fig. 4.18(a).

The field exerts no force on the two arms \(A D\) and \(B C\) of the loop. It is perpendicular to the arm \(A B\) of the loop and exerts a force \(\mathbf{F}_1\) on it which is directed into the plane of the loop. Its magnitude is,
\(
F_1=I b B
\)
Similarly, it exerts a force \(\mathbf{F}_2\) on the arm CD and \(\mathbf{F}_2\) is directed out of the plane of the paper.
\(
F_2=I b B=F_1
\)
Thus, the net force on the loop is zero. There is a torque on the loop due to the pair of forces \(\mathbf{F}_1\) and \(\mathbf{F}_2\). Figure 4.18(b) shows a view of the loop from the AD end. It shows that the torque on the loop tends to rotate it anticlockwise. This torque is (in magnitude),
\(
\begin{aligned}
\tau & =F_1 \frac{a}{2}+F_2 \frac{a}{2} \\
& =I b B \frac{a}{2}+I b B \frac{a}{2}=I(a b) B \\
& =I A B \dots(4.26)
\end{aligned}
\)
where \(A=a b\) is the area of the rectangle.
We next consider the case when the plane of the loop, is not along the magnetic field, but makes an angle with it. We take the angle between the field and the normal to the coil to be angle \(\theta\) (The previous case corresponds to \(\theta=\pi / 2)\). Figure 4.19 illustrates this general case.
The forces on the arms BC and DA are equal, opposite, and act along the axis of the coil, which connects the centres of mass of BC and DA. Being collinear along the axis they cancel each other, resulting in no net force or torque. The forces on arms AB and CD are \(\mathbf{F}_{\mathbf{1}}\) and \(\mathbf{F}_{\mathbf{2}}\). They too are equal and opposite, with magnitude,
\(
F_1=F_2=I b B
\)
But they are not collinear! This results in a couple as before. The torque is, however, less than the earlier case when plane of loop was along the magnetic field. This is because the perpendicular distance between the forces of the couple has decreased. Figure 4.19(b) is a view of the arrangement from the AD end and it illustrates these two forces constituting a couple. The magnitude of the torque on the loop is,
\(
\begin{aligned}
\tau & =F_1 \frac{a}{2} \sin \theta+F_2 \frac{a}{2} \sin \theta \\
& =I a b B \sin \theta \\
& =I A B \sin \theta \dots(4.27)
\end{aligned}
\)

As \(\theta \rightarrow 0\), the perpendicular distance between the forces of the couple also approaches zero. This makes the forces collinear and the net force and torque zero. The torques in Eqs. (4.26) and (4.27) can be expressed as vector product of the magnetic moment of the coil and the magnetic field. We define the magnetic moment of the current loop as,
\(
\mathbf{m}=I \mathbf{A} \dots(4.28)
\)
where the direction of the area vector \(\mathbf{A}\) is given by the right-hand thumb rule and is directed into the plane of the paper in Fig. 4.18. Then as the angle between \(\mathbf{m}\) and \(\mathbf{B}\) is \(\theta\), Eqs. (4.26) and (4.27) can be expressed by one expression
\(
\boldsymbol{\tau}=\mathbf{m} \times \mathbf{B} \dots(4.29)
\)
This is analogous to the electrostatic case (Electric dipole of dipole moment \(\mathbf{p}_{\mathrm{e}}\) in an electric field \(\mathbf{E}\) ).
\(
\boldsymbol{\tau}=\mathbf{p}_{\mathrm{e}} \times \mathbf{E}
\)
As is clear from Eq. (4.28), the dimensions of the magnetic moment are \([\mathrm{A}]\left[\mathrm{L}^2\right]\) and its unit is \(\mathrm{Am}^2\).

From Eq. (4.29), we see that the torque \(\tau\) vanishes when \(m\) is either parallel or antiparallel to the magnetic field \(\mathbf{B}\). This indicates a state of equilibrium as there is no torque on the coil (this also applies to any object with a magnetic moment \(\mathbf{m}\) ). When \(\mathbf{m}\) and \(\mathbf{B}\) are parallel the equilibrium is a stable one. Any small rotation of the coil produces a torque which brings it back to its original position. When they are antiparallel, the equilibrium is unstable as any rotation produces a torque which increases with the amount of rotation. The presence of this torque is also the reason why a small magnet or any magnetic dipole aligns itself with the external magnetic field.

If the loop has \(N\) closely wound turns, the expression for torque, Eq. (4.29), still holds, with
\(
\mathbf{m}=N I \mathbf{A} \dots(4.30)
\)

Example 4.11: A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A . (a) What is the field at the centre of the coil? (b) What is the magnetic moment of this coil?

The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2 T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of \(90^{\circ}\) under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position?
(d) What is the angular speed acquired by the coil when it has rotated by \(90^{\circ}\) ? The moment of inertia of the coil is \(0.1 \mathrm{~kg} \mathrm{~m}^2\).

Solution: (a) From Eq. (4.16)
\(
B=\frac{\mu_0 N I}{2 R}
\)
Here, \(N=100 ; I=3.2 \mathrm{~A}\), and \(R=0.1 \mathrm{~m}\).
Hence,
\(
\begin{aligned}
B & =\frac{4 \pi \times 10^{-7} \times 3.2}{2 \times 10^{-1}}=\frac{4 \times 10^{-5} \times 10}{2 \times 10^{-1}} \quad \text { (using } \pi \times 3.2=10 \text { ) } \\
& =2 \times 10^{-3} \mathrm{~T}
\end{aligned}
\)
The direction is given by the right-hand thumb rule.
(b) The magnetic moment is given by Eq. (4.30),
\(m=N I A=N I \pi r^2=100 \times 3.2 \times 3.14 \times 10^{-2}=10 \mathrm{~A} \mathrm{~m}^2\)
The direction is once again given by the right-hand thumb rule.
(c) \(\tau=|\mathbf{m} \times \mathbf{B}| \quad\) [from Eq. (4.29)]
\(
=m B \sin \theta
\)
Initially, \(\theta=0\). Thus, initial torque \(\tau_i=0\). Finally, \(\theta=\pi / 2\left(\right.\) or \(\left.90^{\circ}\right)\).
Thus, final torque \(\tau_f=m B=10 \times 2=20 \mathrm{~N} \mathrm{~m}\).
(d) From Newton’s second law,
\(g \frac{\mathrm{~d} \omega}{\mathrm{~d} t}=m B \sin \theta\)
where \(g\) is the moment of inertia of the coil. From chain rule,
\(\frac{\mathrm{d} \omega}{\mathrm{~d} t}=\frac{\mathrm{d} \omega}{\mathrm{~d} \theta} \frac{\mathrm{~d} \theta}{\mathrm{~d} t}=\frac{\mathrm{d} \omega}{\mathrm{~d} \theta} \omega\)
Using this,
g \(\omega \mathrm{d} \omega=m B \sin \theta \mathrm{~d} \theta\)
Integrating from \(\theta=0\) to \(\theta=\pi / 2\),
\(
\begin{aligned}
& g \int_0^{\omega f} \omega \mathrm{~d} \omega=m B \int_0^{\pi / 2} \sin \theta \mathrm{~d} \theta \\
& g \frac{\omega_f^2}{2}=-\left.m B \cos \theta\right|_0 ^{\pi / 2}=m B \\
& \omega_f=\left(\frac{2 m B}{g}\right)^{1 / 2}=\left(\frac{2 \times 20}{10^{-1}}\right)^{1 / 2}=20 \mathrm{~s}^{-1}
\end{aligned}
\)

Example 4.12: (a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis).
(b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its orientation of stable equilibrium? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum.
(c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape?

Solution: (a) No, because that would require \(\tau\) to be in the vertical direction. But \(\tau=I \mathbf{A} \times \mathbf{B}\), and since \(\mathbf{A}\) of the horizontal loop is in the vertical direction, \(\tau\) would be in the plane of the loop for any \(\mathbf{B}\).
(b) Orientation of stable equilibrium is one where the area vector \(\mathbf{A}\) of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field.
(c) It assumes circular shape with its plane normal to the field to maximise flux, since for a given perimeter, a circle encloses greater area than any other shape.

4.9.2 Circular current loop as a magnetic dipole

In this section, we shall consider the elementary magnetic element: the current loop. We shall show that the magnetic field (at large distances) due to current in a circular current loop is very similar in behaviour to the electric field of an electric dipole. In Section 4.6, we have evaluated the magnetic field on the axis of a circular loop, of a radius \(R\), carrying a steady current \(I\). The magnitude of this field is [(Eq. (4.15)],
\(
B=\frac{\mu_0 I R^2}{2\left(x^2+R^2\right)^{3 / 2}}
\)
and its direction is along the axis and given by the right-hand thumb rule (Fig. 4.12). Here, \(x\) is the distance along the axis from the centre of the loop. For \(x \gg R\), we may drop the \(R^2\) term in the denominator. Thus,
\(
B=\frac{\mu_0 I R^2}{2 x^3}
\)
Note that the area of the loop \(A=\pi R^2\). Thus,
\(
B=\frac{\mu_0 I A}{2 \pi x^3}
\)
As earlier, we define the magnetic moment \(\mathbf{m}\) to have a magnitude \(I A\), \(\boldsymbol{m}=I \mathbf{A}\). Hence,
\(
\mathrm{B}=\frac{\mu_0 m}{2 \pi x^3}
\)
\(
=\frac{\mu_0}{4 \pi} \frac{2 \mathbf{m}}{x^3} \dots [4.31(\mathrm{a})]
\)
The expression of Eq. [4.31(a)] is very similar to an expression obtained earlier for the electric field of a dipole. The similarity may be seen if we substitute,
\(
\mu_0 \rightarrow 1 / \varepsilon_0
\)
\(\mathbf{m} \rightarrow \mathbf{p}_{\mathrm{e}}\) (electrostatic dipole)
\(\mathbf{B} \rightarrow \mathbf{E} \quad\) (electrostatic field)
We then obtain,
\(
\mathbf{E}=\frac{2 \mathbf{p}_e}{4 \pi \varepsilon_0 x^3}
\)
which is precisely the field for an electric dipole at a point on its axis. considered in Chapter 1, Section 1.10 [Eq. (1.20)].

It can be shown that the above analogy can be carried further. We had found in Chapter 1 that the electric field on the perpendicular bisector of the dipole is given by [See Eq.(1.21)],
\(
E=\frac{\mathbf{p}_e}{4 \pi \varepsilon_0 x^3}
\)
where \(x\) is the distance from the dipole. If we replace \(\mathbf{p} \rightarrow \mathbf{m}\) and \(\mu_0 \rightarrow 1 / \varepsilon_0\) in the above expression, we obtain the result for \(\mathbf{B}\) for a point in the plane of the loop at a distance \(x\) from the centre. For \(x \gg R\),
\(
\mathbf{B}=\frac{\mu_0}{4 \pi} \frac{\mathbf{m}}{x^3} ; \quad \boldsymbol{x} \gg R \dots \text { [4.31(b)] }
\)
The results given by Eqs. [4.31(a)] and [4.31(b)] become exact for a point magnetic dipole.
The results obtained above can be shown to apply to any planar loop: a planar current loop is equivalent to a magnetic dipole of dipole moment \(\mathbf{m}=I \mathbf{A}\), which is the analogue of electric dipole moment \(\mathbf{p}\). Note, however, a fundamental difference: an electric dipole is built up of two elementary units – the charges (or electric monopoles). In magnetism, a magnetic dipole (or a current loop) is the most elementary element. The equivalent of electric charges, i.e., magnetic monopoles, are not known to exist.
We have shown that a current loop (i) produces a magnetic field and behaves like a magnetic dipole at large distances, and (ii) is subject to torque like a magnetic needle. This led Ampere to suggest that all magnetism is due to circulating currents. This seems to be partly true and no magnetic monopoles have been seen so far. However, elementary particles such as an electron or a proton also carry an intrinsic magnetic moment, not accounted by circulating currents.