A loop as a magnetic dipole
A closed loop in a uniform field feels no net force, as lesson 2 showed. But it can still turn. Its behaviour in a field depends on only one vector, fixed by the current, the area and the way the loop faces. A loop that behaves this way is called a magnetic dipole, like a small bar magnet or a compass needle.
For a flat loop of area \(A\) carrying a current \(I\), the magnetic moment is \(\vec M = I\vec A\). The area vector \(\vec A\) has the size of the area and points along the normal to the loop. A coil of \(N\) turns wound together acts as \(N\) loops, so \(\vec M = NI\vec A\). The unit is the ampere metre squared, \(\text{A m}^2\).
Which way along the normal? Curl the fingers of the right hand in the sense of the current round the loop. The thumb points along \(\vec M\). A current running anticlockwise as you look at the loop gives \(\vec M\) pointing towards you.
(a) A coil of \(100\) turns, each of area \(0.02\,\text{m}^2\), carries \(0.5\,\text{A}\). Find its magnetic moment. (b) A rectangular coil of \(40\) turns, \(4\,\text{cm}\) by \(5\,\text{cm}\), carries \(3\,\text{A}\). Find its moment.
(a) \(M = NIA = 100 \times 0.5 \times 0.02 = 1\,\text{A m}^2\), along the normal by the right-hand rule.
(b) \(A = 0.04 \times 0.05 = 2 \times 10^{-3}\,\text{m}^2\), so \(M = 40 \times 3 \times 2 \times 10^{-3} = 0.24\,\text{A m}^2\).
Values used: the numbers given; no constants are needed.
A circular coil of \(20\) turns and radius \(5\,\text{cm}\) carries \(2\,\text{A}\). Find its magnetic moment, in A m².
The couple on a rectangular loop
Take a rectangular loop with sides \(l\) and \(b\), carrying \(I\) in a uniform field \(\vec B\). Let the sides of length \(l\) lie across the field. The other two sides feel forces along the axis of turning, which cancel and do not turn the loop. The two sides of length \(l\) feel forces \(BIl\), in opposite directions, because their currents run opposite ways.
Two equal and opposite forces along different lines form a couple, as in the chapter Torque and Equilibrium. The torque of a couple is one force times the distance between the lines of action. Let \(\theta\) be the angle between \(\vec M\) and \(\vec B\). Then the lines of action are \(b\sin\theta\) apart.
\[ F = BIl \]
Why. each side of length \(l\) lies across the field, so \(\sin 90° = 1\).
\[ \tau = F \times b\sin\theta = BIl\,b\sin\theta \]
Why. the torque of a couple is one force times the distance between the lines of action, which is \(b\sin\theta\) (Fig 4.2).
\[ \tau = BIA\sin\theta = MB\sin\theta \]
Why. \(lb = A\) is the area, and \(IA = M\) for one turn.
\[ \tau = NBIA\sin\theta \]
Why. \(N\) turns wound together carry \(N\) such couples.
\[ \vec\tau = \vec M \times \vec B \]
Read it. the size \(MB\sin\theta\) and the sense, turning \(\vec M\) towards \(\vec B\), are those of the cross product.
The result holds for a flat loop of any shape, not only a rectangle. Any flat loop can be cut into many thin rectangles lying side by side. Neighbouring rectangles share an edge, where their currents run opposite ways and their forces cancel. The torques on the outer edges add up to \(\vec M \times \vec B\) for the whole loop.
A single loop \(0.1\,\text{m}\) by \(0.2\,\text{m}\) carries \(2\,\text{A}\) in \(0.5\,\text{T}\), with its plane along the field. Find (a) the force on each \(0.2\,\text{m}\) side and the torque. A circular coil of \(50\) turns and radius \(0.1\,\text{m}\) carries \(1\,\text{A}\), its moment at \(30°\) to a field of \(0.2\,\text{T}\). Find (b) its torque. Find (c) \(\vec\tau\) for \(\vec M = 2\hat i\,\text{A m}^2\) in \(\vec B = 0.3\hat j\,\text{T}\).
(a) Each \(0.2\,\text{m}\) side lies across the field: \(F = 0.5 \times 2 \times 0.2 = 0.2\,\text{N}\). With the plane along \(\vec B\), \(\theta = 90°\) and the arm is the full \(0.1\,\text{m}\), so \(\tau = 0.2 \times 0.1 = 0.02\,\text{N m}\). Check: \(IAB = 2 \times 0.02 \times 0.5 = 0.02\,\text{N m}\).
(b) \(M = 50 \times 1 \times \pi (0.1)^2 = 0.5\pi = 1.57\,\text{A m}^2\), so \(\tau = 1.57 \times 0.2 \times \sin 30° = 0.157\,\text{N m}\).
(c) \(\vec\tau = 2 \times 0.3\,(\hat i \times \hat j) = 0.6\,\hat k\,\text{N m}\).
A coil with its plane along the field has \(\theta = 90°\) between \(\vec M\) and \(\vec B\). That gives the largest torque, \(NBIA\). A coil with its plane across the field has \(\vec M\) along \(\vec B\) and feels no torque at all. Watch the words: the angle in \(MB\sin\theta\) is between the normal and the field, not between the plane and the field.
A coil of magnetic moment \(0.5\,\text{A m}^2\) sits in a uniform field of \(0.4\,\text{T}\), with its moment at \(30°\) to the field. Find the torque on it, in N m.
A coil has \(\vec M = 3\hat k\,\text{A m}^2\) in a field \(\vec B = 0.2\hat i\,\text{T}\). What is the torque on it?
Where the loop comes to rest
The torque \(MB\sin\theta\) is zero at two angles, \(0°\) and \(180°\). At both, the coil could stay still. But the two rests are very different. Nudge a coil a little from \(\theta = 0°\), and the torque turns it back. Nudge it from \(180°\), and the torque turns it further away.
In Fig 4.3 the coil released at \(150°\), \(90°\) or \(30°\) swings through \(\theta = 0\), overshoots, swings back, and settles with \(\vec M\) along \(\vec B\). The torque readout is largest near \(90°\), at \(0.5\,\text{N m}\), and falls to zero as the coil settles. Released at \(179°\), the coil barely moves at first, because the torque is only \(0.01\,\text{N m}\). Then the small tilt grows, and the coil turns all the way round to \(0°\).
A magnetic dipole in a uniform field feels no torque when its moment is parallel or antiparallel to the field. With \(\vec M\) along \(\vec B\) the equilibrium is stable; with \(\vec M\) opposite to \(\vec B\) it is unstable. The torque is largest, \(MB\), when \(\vec M\) is perpendicular to \(\vec B\).
This is why a compass needle points north. The needle is a small magnetic dipole, and the Earth's field turns it until its moment lies along the field. The chapter Magnetism and Matter takes up the Earth's field and the compass in detail.
A coil hangs free in a uniform field. In which position does it rest in stable equilibrium?
A current loop sits in a uniform magnetic field. Which statements are true?
A single coil in a field would turn to \(\theta = 0\) and stop, as in Fig 4.3. An electric motor must keep turning. So the current in its coil is reversed every half turn, just as the coil passes the stable position. Then the coil's moment points the wrong way, and the torque pushes it on through another half turn. In a mixer-grinder a split ring called a commutator does the switching, pressed by carbon brushes.
- Measuring \(\theta\) from the plane of the loop.In \(\tau = MB\sin\theta\), \(\theta\) is the angle between the normal \(\vec M\) and \(\vec B\). With the plane along \(\vec B\), \(\theta = 90°\).
- Forgetting the number of turns.\(M = NIA\); each turn adds its own couple.
- Taking every zero-torque position as stable.\(\vec M\) along \(\vec B\) is stable; \(\vec M\) against \(\vec B\) is unstable.
- Writing \(\vec B \times \vec M\).The torque is \(\vec M \times \vec B\), which turns \(\vec M\) towards \(\vec B\).