The energy stored by turning
To turn a coil away from \(\vec M \parallel \vec B\), you must push against the torque \(MB\sin\theta\). The work you do is stored. Let the coil go and the field turns it back, giving the energy up as motion. So a dipole in a field has a potential energy that depends on its angle. A raised brick, in the same way, has one that depends on its height.
\[ dW = MB\sin\theta\,d\theta \]
Why. turning the coil slowly through \(d\theta\) against the torque \(MB\sin\theta\) needs this much work.
\[ U(\theta) - U(90°) = \int_{90°}^{\theta} MB\sin\theta\,d\theta = -MB\cos\theta \]
Why. the stored energy is the work done from a chosen reference angle; \(\cos 90° = 0\).
\[ U = -MB\cos\theta = -\vec M \cdot \vec B \]
Read it. take \(U = 0\) at \(\theta = 90°\), the usual choice; only changes in \(U\) matter.
At \(\theta = 0\) the energy is \(-MB = -0.5\,\text{J}\), the bottom of a well. At \(180°\) it is \(+0.5\,\text{J}\), the top of a hump. As in the chapter on Potential Energy and Conservation of Energy, a minimum means stable equilibrium and a maximum unstable. The torque is the slope of the curve. It is steepest, and the torque largest, at \(90°\) and \(270°\).
The electric dipole of the chapter Electric Potential, Energy and the Dipole follows the same rules, with \(\vec p\) for \(\vec M\) and \(\vec E\) for \(\vec B\). The two are twins because both are pairs of equal and opposite forces along different lines.
A coil of \(M = 1\,\text{A m}^2\) sits in \(0.5\,\text{T}\). Find (a) its energy when \(\vec M\) is along \(\vec B\) and when it is opposite. Find (b) the work needed to turn it from along \(\vec B\) to perpendicular, and to opposite. A second coil of \(M = 2\,\text{A m}^2\) sits in \(0.1\,\text{T}\). Find (c) its greatest torque and the energy difference between its two equilibria.
(a) Along: \(U = -MB = -0.5\,\text{J}\). Opposite: \(U = +MB = +0.5\,\text{J}\).
(b) To perpendicular: \(W = U(90°) - U(0°) = 0 - (-0.5) = 0.5\,\text{J}\). To opposite: \(W = 0.5 - (-0.5) = 1\,\text{J}\).
(c) \(\tau_{max} = MB = 2 \times 0.1 = 0.2\,\text{N m}\), at \(90°\). The energy difference is \(2MB = 0.4\,\text{J}\).
Values used: the numbers given; no constants are needed.
A coil of moment \(0.8\,\text{A m}^2\) rests in stable equilibrium in a field of \(0.25\,\text{T}\). How much work is needed to turn it slowly until its moment points against the field, in joules?
On the graph of \(U = -MB\cos\theta\) from \(0°\) to \(360°\), where does the torque on the coil have its greatest size?
Swinging about the stable position
Release a coil at \(90°\) to the field. The torque is at its greatest and turns the coil towards \(\theta = 0\). The coil gains speed, passes \(0\), is slowed by the reversed torque, and stops on the other side. Then it swings back. With no friction it would swing for ever about \(\theta = 0\), like a pendulum. In practice the swing dies away and the coil settles with \(\vec M\) along \(\vec B\).
For small angles, \(\sin\theta \approx \theta\) in radians. The torque is then \(\tau = -MB\theta\), a restoring torque proportional to the angle. In the chapter Dynamics of SHM, a torque \(-\kappa\theta\) on a body of moment of inertia \(I\) gave angular SHM of period \(2\pi\sqrt{I/\kappa}\). Here \(\kappa = MB\).
A coil of moment \(1\,\text{A m}^2\) and moment of inertia \(2 \times 10^{-4}\,\text{kg m}^2\) swings through small angles in a field of \(0.5\,\text{T}\). Find its period.
\(T\) \(= 2\pi\sqrt{I/MB}\) \(= 2\pi\sqrt{2 \times 10^{-4}/0.5}\) \(= 2\pi\sqrt{4 \times 10^{-4}}\) \(= 2\pi \times 0.02\) \(= 0.04\pi\) \(= 0.126\,\text{s}\).
A stronger field or a larger moment gives a stiffer "spring" and a shorter period.
A coil of moment \(2\,\text{A m}^2\) and moment of inertia \(5 \times 10^{-4}\,\text{kg m}^2\) swings through small angles in a field of \(0.1\,\text{T}\). Find its period, in seconds.
An electron in orbit is a current loop
A charge that goes round and round a circle makes a current. Stand at one point of the path. A charge \(e\) moving at speed \(v\) round a circle of radius \(r\) passes you \(v/2\pi r\) times each second. So the charge passing each second, the current, is \(I = ev/2\pi r\).
The current runs round an area \(\pi r^2\). So the orbit is a current loop with moment \(M = IA = (ev/2\pi r)(\pi r^2) = evr/2\). The old picture of the hydrogen atom has the electron on just such an orbit. Unit 16, Modern Physics, takes up that atom; here we need only its moving charge.
Fig 5.1 shows the directions. The electron's angular momentum is \(L = m_evr\), out of the page for anticlockwise motion. But the electron is negative, so the current runs the other way, clockwise. The right-hand rule then puts \(\vec M\) into the page, opposite to \(\vec L\). Divide the sizes: \(M/L = (evr/2)/(m_evr) = e/2m_e\).
In the old model of the hydrogen atom, the electron circles at \(v = 2.2 \times 10^6\,\text{m/s}\) on a radius \(r = 5.3 \times 10^{-11}\,\text{m}\). Find (a) the current, (b) the magnetic moment, (c) the angular momentum and the ratio \(M/L\).
(a) \(I\) \(= ev/2\pi r\) \(= 1.6 \times 10^{-19} \times 2.2 \times 10^6/(2\pi \times 5.3 \times 10^{-11})\) \(= 1.06 \times 10^{-3}\,\text{A}\), about a milliampere.
(b) \(M\) \(= evr/2\) \(= 1.6 \times 10^{-19} \times 2.2 \times 10^6 \times 5.3 \times 10^{-11}/2\) \(= 9.3 \times 10^{-24}\,\text{A m}^2\).
(c) \(L\) \(= m_evr\) \(= 9.1 \times 10^{-31} \times 2.2 \times 10^6 \times 5.3 \times 10^{-11}\) \(= 1.06 \times 10^{-34}\,\text{kg m}^2\text{/s}\), and \(M/L = e/2m_e = 8.8 \times 10^{10}\,\text{C/kg}\).
Values used: \(e = 1.6 \times 10^{-19}\,\text{C}\), \(m_e = 9.1 \times 10^{-31}\,\text{kg}\).
An electron moves at \(1.1 \times 10^6\,\text{m/s}\) round a circle of radius \(2.12 \times 10^{-10}\,\text{m}\). Treat it as a current loop and find its magnetic moment, in units of \(10^{-23}\,\text{A m}^2\).
The ratio of magnetic moment to angular momentum is called the gyromagnetic ratio. For any body whose charge and mass are spread in the same proportion, it is \(q/2m\), whatever the shape or the speed of spin. So a uniformly charged sphere of charge \(Q\), mass \(m\) and radius \(R\), spinning at \(\omega\), has \(M = (Q/2m) \times \tfrac25 mR^2\omega = \tfrac15 QR^2\omega\). The electron's own spin breaks this rule: its ratio is almost exactly twice \(e/2m_e\).
An MRI scanner, like those at AIIMS in New Delhi, holds the patient in a steady field of about \(1.5\,\text{T}\). Every hydrogen nucleus in the body is a tiny magnetic dipole, and the field lines them up. Lying along the field is the state of lower energy \(-MB\). A pulse of radio waves, near \(64\,\text{MHz}\) at \(1.5\,\text{T}\), tips the dipoles away. As they swing back, they send out faint radio signals, and a computer builds a picture of the soft tissue from them.
- Putting the energy minimum at 90°.\(U = -MB\cos\theta\) is least at \(0°\), where \(\vec M\) lies along \(\vec B\); it is zero at \(90°\).
- Using \(\sin\theta\) in the energy.The torque has \(\sin\theta\); the energy has \(\cos\theta\) and a minus sign.
- Giving an orbiting electron's moment the direction of its motion.The electron is negative, so its current and its moment point opposite to what its motion suggests; \(\vec M\) is opposite to \(\vec L\).