Chapter 8
Sources of Magnetic Field

8.1.1  Magnetic Field for a Single Moving Charge

In keeping with our previous statement about the association of charges acted upon by fields and the fields produced by those charges, it is only natural to think that, once it is established that magnetic fields act on moving electric charges, then moving electric charges should create magnetic fields. In fact, experiment shows just such a relationship! The symmetry inherent in nature between fields and charges is therefore upheld. We find that a single moving point charge q which has constant velocity v in a vacuum produces a magnetic field

® B   = m0 4p q ® v   × ^ r   r2

(8.1.1.1)

Note the introduction of a new fundamental constant of nature, m₀, the permeability of free space. We define the value of m₀ as being

m0 = 4p×10-7 N·s2/C2 = 4p×10-7 T·m/A

(8.1.1.2)

So, if m₀ is a constant of nature, why is it defined rather than measured as other constants have been? The reason lies in what was historically recognized as a curious relationship between the permittivity and the permeability of free space. Namely, their product is the inverse of the square of the speed of light, i.e.

c2 = 1 e0m0 .

(8.1.1.3)

The reason for this amazing correspondence will be explained later in the semester. For now, we take it that it is useful to define the value of c and hence after measuring e₀ we define the value of m₀ to be consistent with the definition of c given the above correspondence.

8.1.2  Biot-Savart Law

Figure 8.1: Magnetic field around a straight conductor carrying a current I.

Starting from experimental observation of the above, a theoretical formula can be obtained that reflects the contribution due to current through infinitesimal lengths of wire all summed together vectorially through integration. This formula, first worked out by Biot and Savart, is referred to as the Biot-Savart Law

d ® B   = m0 4p I d ® l   × ^ r   r2 = m0I 4p d ® l   × ® r   r3

(8.1.2.4)

is the magnetic field contribution of a current element, i.e. an infinitesimal length dl of current I where dl has the direction of I and r is the vector from the current element to the position where the field is to be calculated. Note that implicit in the Biot-Savart Law is the notion that magnetic fields obey the principle of superposition. This means that we can get the net field by just vectorially adding the contributions of the fields due to the infinite number of infinitesimal current elements.

8.1.3  Applying the Biot-Savart Law

To gain experience with applying this law, let's consider some specific examples, starting with the infinite straight wire carrying a steady current I.

Magnetic Field of a Long, Thin Wire

Problem 1:

Find the magnetic field at a distance R from a long, thin wire carrying a current I as shown in figure  8.2.

Figure 8.2: An infinitely long wire carrying a current I and the magnetic field a radial distance R from the wire.

Solution:

We first note that the direction of the B field, as given by the Biot-Savart Law, is into the page. A modified right-hand rule allows us to find the direction of the field more succintly: point the thumb of your right hand along the direction of the current and your fingers naturally curl in the direction of the B field. To get the magnitude for the field, we need to integrate over the infinite length of the wire. For the diagram shown, the location of dl is a distance l from the point directly across from the point P where we wish to find the B field. Therefore,

r = Ö   R2 + l2

(8.1.2.5)

and

|d ® l   × ® r   | = r dl sinq = R dl

(8.1.2.6)

So, the integration is straightforward as we note that, by symmetry, the integration from 0 to infinity must yield the same result as the integration from 0 to negative infinity, thus

B = 2 ó õ ¥ 0  m0I 4p R dl (l2 + R2)3/2 = m0I 2p é ê ê ê ë l Ö l2 +R2 ù ú ú ú û ¥ 0  B = m0I 2pR (8.1.2.7)

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Magnetic Field of a Wire Loop

Problem 2:

Assume that the wire in figure  has a current I through it in the direction shown. What is the magnitude of the magnetic field produced at point C due to

1. the straight segments
2. the semicircular arc
3. the entire wire

Figure 8.3: Calculating the B contributions at point C due to the straight and curved sections of the wire.

Solution:

1. Looking at the figure, we see that the straight sections have infinitesimal lengths that lie parallel or antiparallel to r, the vector that goes from dl to point C. Therefore, dl×r = 0 in both cases for every dl along the straight lengths.
2. For the semicircular arc, each infinitesimal segment, dl, is an equal distance R  from point C. For each such segment, dl×r is directed into the page and the magnitude is R dl  since ds and r  are perpendicular at each point around the arc. Therefore, by the Biot-Savart Law, we have
   

   Barc = ó õ pR 0  m0I 4p R dl R3 = m0I 4pR2 [l]0pR Barc = m0I 4R (8.1.2.8)

3. To get the net contribution, just add the results for parts a. and b. of the question
   

   Bnet = Barc = m0I 4R

   (8.1.2.9)

   The net field points into the page as described in part b.

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Magnetic Field of Multiple Arcs

Problem 3:

Consider the circuit shown in figure 8.4. A current I runs through it. Given the parameters shown in the figure, find the magnetic field at point P.

Figure 8.4: Find the magnetic field at point P for this current distribution.

Solution:

We can take our result from the previous problem and apply it here directly. The straight sections yield no contribution to the field at P since they lie along the radial line from each current element to point P. We also know that a semicircular arc gives a magnetic field of magnitude

B = m0I 4R

(8.1.2.10)

An arc which is less than a semicircle gives a field of

Barc = m0I 4R arc-length pR

(8.1.2.11)

i.e. we find the fraction of a semicircle which the arc represents. For our problem, we use the right-hand rule to note that the field due to the arc of radius b points out of the page while the arc of radius a gives a field that points into the page. If we arbitrarily define into the page as negative, then

Bnet = Bb + Ba = m0I 4b bq pb - m0I 4a aq pa Bnet = m0I(a - b)q 4pab (8.1.2.12)

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Magnetic Field of a Hairpin Loop

Problem 4:

Find the magnetic field at point P for figure  8.5 which shows a current I traveling a long, straight wire length into a semicircular wire of radius R and out along another long, straight section.

Figure 8.5: Find the magnetic field at point P for the current distribution shown.

Solution:

Let's consider the current distribution as consisting of 3 parts as labeled in figure  8.6.

Figure 8.6: Contributions to the magnetic field at point P  due to the straight and curved sections of the current distribution.

These pieces are the two long, straight sections and the semicircular arc. For pieces 1 and 3, note that their contributions to the net B field at point P are both out of the page. In setting up the Biot-Savart integral for infinitesimal lengths of current elements dx, note that the setup of the integral will be identical for both, namely, we can characterize the position of dx in both cases w.r.t. the places where the straight sections combine with the semi-circular arcs, hence

B1 + B3 = 2 m0I 4p ó õ ¥ 0  r sinq r3 dx = m0I 2p ó õ ¥ 0  R r3 dr = m0IR 2p ó õ ¥ 0  1 (R2 + x2)3/2 dx = m0IR 2p x R2   ______ ÖR2 + x2   ê ê ê ê ê ¥ 0  B1 + B3 = m0I 2pR (8.1.2.13)

Therefore, the sum of these two contributions is the same as that of a single infinite length line carrying a current I. To get the contribution from piece 2, the semi-circular arc, look again at figure 8.6. In this case, each current element is a distance R away from point P and the angle between ds, the infinitesimal arc-length of current, and r, the vector from ds to point P, is 90^°  because ds is tangential to the arc and r is along the radial direction. So,

B2 = m0I 4p ó õ pR 0  r ds r3 = m0I 4p ó õ pR 0  ds R2 = m0I 4pR2 (pR) B2 = m0I 4R (8.1.2.14)

Summing all the contributions together yields

Bnet = B1 + B2 + B3 = m0I R é ê ë 1 2p + 1 4 ù ú û

(8.1.2.15)

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