Adding the pieces of a bent wire
A bent wire is a chain of short straight elements. Each feels \(d\vec F = I\,d\vec l \times \vec B\). In a uniform field \(\vec B\) is the same vector at every element, and so is \(I\). Both can come out of the sum: \(\vec F = I\left(\sum d\vec l\right) \times \vec B\). Only the sum of the little length vectors is left to find.
Place the length vectors head to tail along the wire. Their sum is the single straight vector from the start of the wire to its end, whatever the path in between. We call it the effective length, \(\vec L_{eff}\). The shape of the wire has dropped out of the answer.
A wire of any shape carries a current \(I\) from a point A to a point B in a uniform magnetic field. It feels the force \(\vec F = I\vec L_{eff} \times \vec B\), where \(\vec L_{eff}\) is the straight vector from A to B. It feels the same force as a straight wire from A to B.
Look at the semicircle in Fig 2.1. Near A and B its pieces run up and down, so their forces point sideways. The forces near A point left and those near B point right, so they cancel. Only the upward parts are left, and they add to the force on the straight diameter. The zigzag is longer still, but its extra sideways pushes cancel in the same way.
A field of \(0.2\,\text{T}\) points into the page. A semicircular wire of radius \(0.1\,\text{m}\) carries \(5\,\text{A}\) from A to B, the ends of its diameter. Find (a) the force on it. Another wire runs \(0.3\,\text{m}\) straight, then round a semicircle of radius \(0.1\,\text{m}\), then \(0.3\,\text{m}\) on, all along one line. It carries \(5\,\text{A}\). Find (b) the force on it.
(a) \(L_{eff}\) is the diameter, \(2 \times 0.1 = 0.2\,\text{m}\). So \(F = BIL_{eff} = 0.2 \times 5 \times 0.2 = 0.2\,\text{N}\), perpendicular to AB in the plane of the page.
(b) The ends lie \(0.3 + 0.2 + 0.3 = 0.8\,\text{m}\) apart on a straight line, so \(L_{eff} = 0.8\,\text{m}\) and \(F = 0.2 \times 5 \times 0.8 = 0.8\,\text{N}\).
Values used: the numbers given; no constants are needed.
A wire runs \(0.3\,\text{m}\) along the \(x\) axis and then \(0.4\,\text{m}\) along the \(y\) axis. It carries \(2\,\text{A}\) in a uniform field of \(0.5\,\text{T}\) along the \(z\) axis. Find the size of the total force, in newtons.
A closed loop
Now close the wire into a loop. Its start and its end are the same point, so \(\vec L_{eff} = 0\). A uniform field then exerts no net force on the loop, whatever its shape. The pieces are still pushed, often hard, but the pushes add to zero.
In Fig 2.2 each side feels \(F = BIL = 2.0 \times 5 \times 0.1 = 1.0\,\text{N}\). The three forces are equal and point at \(120°\) to one another, so they cancel. Squeezed from three sides, the loop stays where it is. Reverse the current and all three forces point outward; they still cancel.
"No net force" is not "no effect". Lay a square loop flat in a field that lies in its plane. The two sides across the field are pushed, one out of the page and one into it. These equal and opposite forces act along different lines. So they form a pair that turns the loop, as lesson 4 of this chapter shows.
A square loop of side \(0.2\,\text{m}\) carries \(4\,\text{A}\). A uniform field of \(0.5\,\text{T}\) lies in the plane of the loop, parallel to two of its sides. What does the loop feel?
Explain, using the effective-length rule, why a uniform magnetic field exerts no net force on a current loop of any shape.
In a uniform field every element feels \(I\,d\vec l \times \vec B\) with the same \(I\) and \(\vec B\), so the total is \(I\left(\sum d\vec l\right) \times \vec B\). The length vectors, placed head to tail around the loop, end where they began, so their sum is zero. A zero vector crossed with \(\vec B\) is zero, so the net force vanishes for any shape of loop.
Adding the forces round an arc
The effective-length rule is a shortcut. It is worth checking it once by adding the forces directly. Take a semicircle of radius \(R\) above its diameter, carrying \(I\) in a field \(B\) into the page. Mark each element by its angle \(\theta\) from the diameter. The element has length \(R\,d\theta\) and lies across the field, so its force is \(BIR\,d\theta\), pointing straight out from the centre.
\[ dF = BIR\,d\theta \]
Why. the element \(R\,d\theta\) lies across the field, so \(dF = BI\,(R\,d\theta)\), directed straight out from the centre.
\[ dF_x = BIR\cos\theta\,d\theta, \quad dF_y = BIR\sin\theta\,d\theta \]
Why. resolve the outward force along the diameter (\(x\)) and along the perpendicular bisector (\(y\)).
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Your turn. the \(x\) parts on the two halves cancel in pairs. Which line is it?
\[ F = 2BIR \]
Read it. the force points along the perpendicular bisector of the diameter, the same as for a straight wire of length \(2R\).
The answer, \(2BIR\), is \(BI\) times the diameter. So the direct sum agrees with the effective-length rule. For \(R = 0.1\,\text{m}\), \(I = 5\,\text{A}\) and \(B = 0.2\,\text{T}\), it gives \(0.2\,\text{N}\), as in Fig 2.1. The direct method is needed when the field changes from place to place, where the shortcut fails.
Turn the same semicircle so that the field lies in its plane, perpendicular to the diameter AB. \(\vec L_{eff}\) is still AB, across the field, so the force is still \(BI \times 2R = 0.2\,\text{N}\). But now it points out of the plane of the semicircle. If instead the field lies along AB, then \(\vec L_{eff} \parallel \vec B\) and the force is zero.
A semicircular wire joins the ends A and B of its diameter and carries a current. A uniform field lies in the plane of the semicircle, parallel to AB. What is the net force on the wire?
When the field is not uniform
The shortcut took \(\vec B\) out of the sum because it was the same at every element. If \(\vec B\) changes along the wire, it cannot come out. Then each part of the wire must be treated with the field where that part sits. A closed loop can now feel a net force.
In the growing field of Fig 2.3 the left side, at \(x = 0\), sits in \(B_0 = 0.1\,\text{T}\). It feels \(IB_0l = 5 \times 0.1 \times 0.2 = 0.1\,\text{N}\) to the left. The right side, at \(x = l\), sits in \(2B_0\) and feels \(0.2\,\text{N}\) to the right. The top and bottom span the same range of \(x\) with opposite currents, so their forces still cancel. The net force is \(IB_0l = 0.1\,\text{N}\), towards the stronger field.
The effective-length rule and "a loop feels no net force" both need a uniform field. In a field that changes from place to place, treat each part of the wire in its own field. Then add the forces.
A square loop of side \(0.1\,\text{m}\) carries \(2\,\text{A}\) in a field \(B = 0.3(1 + x/l)\,\text{T}\) at right angles to its plane, where \(l\) is the side. Two sides lie along \(x = 0\) and \(x = l\). Find the size of the net force on the loop, in newtons.
An electromagnetic launcher is the rod on the rails of lesson 1 made enormous. A huge current flows up one rail, across a sliding armature and down the other. The current in the rails makes a strong field between them. The field pushes the armature along with a force \(BIL\), faster and faster, until it leaves the end at high speed. Navies have tested such launchers, and the same idea may one day send small loads towards orbit.
- Using the length of the wire instead of the distance between its ends.In a uniform field only \(\vec L_{eff}\), the straight vector from start to end, enters.
- Saying a loop in a uniform field feels nothing.The net force is zero, but the pushes on opposite sides can still turn the loop.
- Adding the sizes of the element forces round an arc.Resolve each force into components first; the components across the bisector cancel.
- Using the effective-length rule in a non-uniform field.Find the force on each part with its own field, then add.