Chapter 7
Magnetic Fields and Forces

As with the case of the electric field, we hypothesize the existence of such a field as a way of explaining action at a distance in which objects exert forces on each other without direct mechanical contact. As it is easy to distinguish between gravitational and electrical forces (and hence the nature of the field produced was originally hypothesized as being quite different in nature: electric fields act only on electric charges, gravitational fields produce forces only on mass, etc.), so too were magnetic fields and electric fields thought to be very different. After all, the nature of the forces produced are very different as we will see. However, there is a very important reason for thinking, before we know very much about magnetic fields, that they are closely linked with electric fields. Although this could not be known before the atomic theory of matter was believed and the basic constituents of atoms discovered, magnetic forces act only on electric charges as there are no confirmed observations of magnetic charges. The fact that magnetic fields exert forces on electric charges at all is a clue that somehow electricity and magnetism share a deep bond. To understand that bond, we need to explore thoroughly the properties of the magnetic field.

7.1.1  Magnetic Force on a Charged Particle

Experimentally, to isolate the nature of the magnetic field, we need to start, as with the electric field, with the simplest thing you can act on, namely a particle. The behavior of magnetic forces is best indicated by how a charged particle behaves in the presence of such a field (we will deal with the question of how the field is produced in the first place later). To see this, explore with the Java applet below.

In the applet, a charged particle moves in the presence of a magnetic field that is perpendicular to the page. If the particle stops its motion, the magnetic force on it is zero. This is true independent of the direction or magnitude of the magnetic field. The mathematical description of the force due to the magnetic field is

® F   = q ® v   × ® B

(7.1.1.1)

This formula describes a force which is, by virtue of the cross-product, always perpendicular to v and B and has a magnitude equal to qvB sinq where q is the angle between v and B. This is a general result for any charged particle moving in any magnetic field.

The circular path seen in the applet results because the force is always perpendicular to the velocity (and also to the B field direction). If the velocity magnitude and B are constant, then the path of the particle is a circle since the force, and hence the acceleration, of the particle is always perpendicular to its path and constant in magnitude. In this case, we satisfy the centripetal condition in that,

| ® F   B  | = mv2 r qvB = mv2/r Þ r = mv qB (7.1.1.2)

This fact is used to make two very practical devices of the 20th century: the mass spectrometer and the cyclotron. Both of these have a multitude of scientific and practical uses.

7.1.2  Forces on Currents

In the earliest studies of magnetic fields, individual charged particles were not known. The properties of magnetic fields and forces were studied using the effects of fields on current distributions. If we have a bunch of charges flowing through a straight wire of length L, then the net force is

Figure 7.2: Force on a wire carrying a current I in a magnetic field B.

where the direction of L is the same as the direction of the current.

We can apply the above formulas to the more general case of current distributions of arbitrary shape by breaking any line of current into infinitesimal lengths and integrating over them to find the net force,

d ® F   = i d ® L   × ® B   .

(7.1.2.3)

Let's try this idea out on the following example.

Problem:

Find the force on the current distribution shown below in a magnetic field B. Assume all quantities shown in the figure are given.

Figure 7.3: A semi-circular loop and two straight wire lengths carrying a current I in a magnetic field B  pointing into the page.

Solution:

We break the current into three pieces as shown in the figure below.

Figure 7.4: The forces on each part of the wire.

The two straight pieces are identical as far as how the magnetic field interacts with them to produce a force. These forces have directions as shown in the figure due to the right-hand rule. Their magnitudes are

F1 = F3 = iLB

(7.1.2.4)

For the loop, we break it into infinitesimal pieces of arc-length ds as shown below.

Figure 7.5: Calculating the force on the loop.

As can be seen from the figure, if we choose symmetrical positions for ds above and below the midpoint of the semicircular loop, we find that the y components of the force produced cancel while the x components add. Thus, for each infinitesimal length ds we wish to find

dFnet = dFx = i ds Bsinq

(7.1.2.5)

Since the B field is at right angles to the current at every point on the current distribution. To get ds in terms of theta, note that ds = R dq. We can finish the calculation from here.

F2, net = Fy = ó õ p 0  iRBsinq dq = iRB[-cosq]0p F2, net = 2iRB (7.1.2.6)

The total force is therefore to the left with magnitude

Fnet = F1 + F2 + F3 = 2iB(R + L)

(7.1.2.7)

7.1.3  Magnetic Force on an Arbitrarily Shaped Wire

We can generalize the above result to take into account any arbitrary shape of current distribution. For example, consider the following figure where some arbitrarily shaped wire takes a current i from point a to point b in a magnetic field B which has constant magnitude and direction perpendicular to the plane containing i.

Figure 7.6: An arbitrarily shaped current distribution in an external magnetic field.

The problem is not as difficult as it might at first appear. Consider any particular piece of the wire.

Figure 7.7: Blowup of a section of the wire.

In breaking this distribution into infinitesimal lengths, dl, if we note that we are free to do this such that dl has components that are both perpendicular and parallel to the line between points a and b (we called these the y and x axes, respectively), then note that the contributions on the force due to the y components will cancel. Why? Note that in going from a to b, our net displacement is only along x (remember: we have chosen the x axis to lie along the line between a and b). Hence, any +y displacement must eventually be countered with a -y displacement that cancels it. Since the magnetic force on any +y displacement is opposite in direction to that on any -y displacement, the integral over i*dl_yB must be zero. Hence, we have a non-zero contribution only from the i*dl_xB components. We note that the B field is perpendicular to both the x and y components. Hence our result is

Fnet = iLabB

(7.1.3.8)

where L_ab represents the distance from a to b along a straight line between the two points. If you go back to our previous example, you will see that this result is exactly consistent with this prediction.

Send comments to larryg@upenn5.hep.upenn.edu.
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