29.4 The Magnetic Field of a Current

Moving charges are the source of magnetic fields, but the magnetic fields of current-carrying wires—immense numbers of charges moving together—are much more important than the feeble magnetic fields of individual charges. Real current-carrying wires, with their twists and turns, have very complex fields. However, we can once again focus on the physics by using simplified models. It turns out that three common magnetic field models are the basis for understanding a wide variety of magnetic phenomena. We present them here together as a reference; the next few sections of this chapter will be devoted to justifying and explaining these results.

MODEL 29.1 Three key magnetic fields

An infinite wire:

Straight wires

The figure illustrates a current-carrying horizontal wire placed in a magnetic field.

Full Alternative Text

B = \frac{μ_{0}}{2 π} \frac{I}{r}

A current loop:

Flat coils

The figure illustrates side view of flat coil with N turns in a magnetic field.

Full Alternative Text

B_{center} = \frac{μ_{0}}{2} \frac{N I}{R}

A solenoid:

Helical coils

The figure illustrates a current carrying helical coil in a uniform magnetic field.

Full Alternative Text

To begin, we need to write the Biot-Savart law in terms of current. FIGURE 29.13a shows a current-carrying wire. The wire as a whole is electrically neutral, but current I represents the motion of positive charge carriers through the wire. Suppose the small amount of moving charge $Δ q$ spans the small length $Δ s$ . The charge has velocity $\vec{v} = Δ \vec{s} / Δ t$ , where the vector $Δ \vec{s}$ , which is parallel to $\vec{v}$ , is the charge’s displacement vector. If $Δ q$ is small enough to treat as a point charge, the magnetic field it creates at a point in space is proportional to $(Δ q) \vec{v}$ . We can write $(Δ q) \vec{v}$ in terms of the wire’s current I as

(Δ q) \vec{v} = Δ q \frac{Δ \vec{s}}{Δ t} = \frac{Δ q}{Δ t} Δ \vec{s} = I Δ \vec{s}

(29.5)

where we used the definition of current, $I = Δ q / Δ t$ .

The figure labeled ("a") illustrates a current-carrying wire in motion.

Full Alternative Text

The figure labeled (b) illustrates a moving current-carrying wire in a plane.

Full Alternative Text

If we replace $q \vec{v}$ in the Biot-Savart law with $I Δ \vec{s}$ , we find that the magnetic field of a very short segment of wire carrying current I is

\begin{matrix} \vec{B}_{current segment} = \frac{μ_{0}}{4 π} \frac{I Δ \vec{s} \times \hat{r}}{r^{2}} \\ (magnetic field of a very short segment of current) \end{matrix}

(29.6)

Equation 29.6 is still the Biot-Savart law, only now written in terms of current rather than the motion of an individual charge. FIGURE 29.13b shows the direction of the current segment’s magnetic field as determined by using the right-hand rule.

Equation 29.6 is the basis of a strategy for calculating the magnetic field of a current-carrying wire. You will recognize that it is the same basic strategy you learned for calculating the electric field of a continuous distribution of charge. The goal is to break a problem down into small steps that are individually manageable.

PROBLEM-SOLVING STRATEGY 29.1
The magnetic field of a current

MODEL Model the wire as a simple shape.

VISUALIZE For the pictorial representation:

Draw a picture, establish a coordinate system, and identify the point P at which you want to calculate the magnetic field.
Divide the current-carrying wire into small segments for which you already know how to determine $\vec{B}$ . This is usually, though not always, a division into very short segments of length $Δ s$ .
Draw the magnetic field vector for one or two segments. This will help you identify distances and angles that need to be calculated.

SOLVE The mathematical representation is $\vec{B}_{net} = \sum \vec{B}_{i}$ .

Write an algebraic expression for each of the three components of $\vec{B}$ (unless you are sure one or more is zero) at point P. Let the $(x, y, z)$ -coordinates of the point remain as variables.
Express all angles and distances in terms of the coordinates.
Let $Δ s \to d s$ and the sum become an integral. Think carefully about the integration limits for this variable; they will depend on the boundaries of the wire and on the coordinate system you have chosen to use.

REVIEW Check that your result is consistent with any limits for which you know what the field should be.

The key idea here, as it was in Chapter 23, is that integration is summation. We need to add up the magnetic field contributions of a vast number of current segments, and we’ll do that by letting the sum become an integral.

EXAMPLE 29.3 The magnetic field of a long, straight wire

A long, straight wire carries current I in the positive x-direction. Find the magnetic field at distance r from the wire.

MODEL Model the wire as being infinitely long.

VISUALIZE FIGURE 29.14 illustrates the steps in the problem-solving strategy. We’ve chosen a coordinate system with point P on the y-axis. We’ve then divided the wire into small segments, labeled with index i, each containing a small amount $Δ q$ of moving charge. Unit vector $\hat{r}$ and angle $θ_{i}$ are shown for segment i. You should use the right-hand rule to convince yourself that $\vec{B}_{i}$ points out of the figure, in the positive z-direction. This is the direction no matter where segment i happens to be along the x-axis. Consequently, $B_{x}$ (the component of $\vec{B}$ parallel to the wire) and $B_{y}$ (the component of $\vec{B}$ straight away from the wire) are zero. The only component of $\vec{B}$ we need to evaluate is $B_{z}$ , the component tangent to a circle around the wire.

The figure illustrates the method of calculating the magnetic field of a long, straight wire carrying current I.

FIGURE 29.14 Full Alternative Text

SOLVE We can use the Biot-Savart law to find the field ${(B_{i})}_{z}$ of segment i. The cross product $Δ \vec{s}_{i} \times \hat{r}$ has magnitude $(Δ x) (1) sin θ_{i}$ , hence

{(B_{i})}_{z} = \frac{μ_{0}}{4 π} \frac{I Δ x sin θ_{i}}{r_{i}^{2}} = \frac{μ_{0}}{4 π} \frac{I sin θ_{i}}{r_{i}^{2}} Δ x = \frac{μ_{0}}{4 π} \frac{I sin θ_{i}}{x_{i}^{2} + y^{2}} Δ x

where we wrote the distance $r_{i}$ in terms of $x_{i}$ and y. We also need to express $θ_{i}$ in terms of $x_{i}$ and y. Because $sin (180 ° - θ) = sin θ$ , this is

sin θ_{i} = sin (180 ° - θ_{i}) = \frac{y}{r_{i}} = \frac{y}{\sqrt{x_{i}^{2} + y^{2}}}

With this expression for $sin θ_{i}$ , the magnetic field of segment i is

{(B_{i})}_{z} = \frac{μ_{0}}{4 π} \frac{I y}{{(x_{i}^{2} + y^{2})}^{3 / 2}} Δ x

Now we’re ready to sum the magnetic fields of all the segments. The superposition is a vector sum, but in this case only the z-components are nonzero. Summing all the ${(B_{i})}_{z}$ gives

B_{wire} = \frac{μ_{0} I y}{4 π} \sum_{i}^{} \frac{Δ x}{{(x_{i}^{2} + y^{2})}^{3 / 2}} \to \frac{μ_{0} I y}{4 π} \int_{- \infty}^{\infty} \frac{d x}{{(x^{2} + y^{2})}^{3 / 2}}

Only at the very last step did we convert the sum to an integral. Then our model of the wire as being infinitely long sets the integration limits at $\pm \infty$ . This is a standard integral that can be found in Appendix A or with integration software. Evaluation gives

B_{wire} = \frac{μ_{0} I y}{4 π} \frac{x}{y^{2} {(x^{2} + y^{2})}^{1 / 2}} |_{- \infty}^{\infty} = \frac{μ_{0}}{2 π} \frac{I}{y}

This is the magnitude of the field. The field direction is determined by using the right-hand rule.

The coordinate system was our choice, and there’s nothing special about the y-axis. The symbol y is simply the distance from the wire, which is better represented by r. With this change, the magnetic field is

{\vec{B}}_{wire} = (\frac{μ_{0}}{2 π} \frac{I}{r}, \begin{array}{l} tangent to a circle around the wire \\ in the right-hand direction \end{array})

REVIEW FIGURE 29.15 shows the magnetic field of a current-carrying wire. Compare this to Figure 29.2 and convince yourself that the direction shown agrees with the right-hand rule.

The figure illustrates the magnetic field of a long, straight wire carrying a current labeled I.

FIGURE 29.15 Full Alternative Text

The magnetic field of an infinite, straight wire was the first of our key magnetic field models. Example 29.3 has shown that the field has magnitude

B_{wire} = \frac{μ_{0}}{2 π} \frac{I}{r} (long, straight wire)

(29.7)

The field direction, encircling the wire, is given by the right-hand rule.

EXAMPLE 29.4 The magnetic field strength near a heater wire

A 1.0-m-long, 1.0-mm-diameter nichrome heater wire is connected to a 12 V battery. What is the magnetic field strength 1.0 cm away from the wire?

MODEL 1 cm is much less than the 1 m length of the wire, so we will model the wire as infinitely long.

SOLVE The current through the wire is $I = Δ V_{bat} / R$ , where the wire’s resistance R is

R = \frac{ρ L}{A} = \frac{ρ L}{π r^{2}} = 1.9 Ω

The nichrome resistivity $ρ = 1.50 \times 10^{- 6} Ω m$ was taken from Table 27.2. Thus the current is $I = (12 V) / (1.9 Ω) = 6.3 A$ . The magnetic field strength at distance $d = 1.0 cm = 0.010 m$ from the wire is

\begin{matrix} B_{wire} & = & \frac{μ_{0}}{2 π} \frac{I}{d} = \frac{(1.26 \times 10^{- 6} T m / A) (6.3 A)}{2 π (0.010 m)} \\ = & 1.3 \times 10^{- 4} T \end{matrix}

REVIEW The magnetic field of the wire is slightly more than twice the strength of the earth’s magnetic field.

Motors, loudspeakers, metal detectors, and many other devices generate magnetic fields with coils of wire. The simplest coil is a single-turn circular loop of wire. A circular loop of wire with a circulating current is called a current loop.

EXAMPLE 29.5 The magnetic field of a current loop

FIGURE 29.16a shows a current loop, a circular loop of wire with radius R that carries current I. Find the magnetic field of the current loop at distance z on the axis of the loop.

The figure labeled "(a)" illustrates a practical current loop.

The figure labeled (b) illustrates an ideal current loop.

Full Alternative Text

MODEL Real coils need wires to bring the current in and out, but we’ll model the coil as a current moving around the full circle shown in FIGURE 29.16b.

VISUALIZE FIGURE 29.17 shows a loop for which we’ve assumed that the current is circulating ccw. We’ve chosen a coordinate system in which the loop lies at $z = 0$ in the xy-plane. Let segment i be the segment at the top of the loop. Vector $Δ \vec{s}_{i}$ is parallel to the x-axis and unit vector $\hat{r}$ is in the yz-plane, thus angle $θ_{i}$ , the angle between $Δ \vec{s}_{i}$ and $\hat{r}$ , is $90 °$ .

The figure illustrates the method of calculating the magnetic field of a current loop.

FIGURE 29.17 Full Alternative Text

The direction of $\vec{B}_{i}$ , the magnetic field due to the current in segment i, is given by the cross product $Δ \vec{s}_{i} \times \hat{r}$ . $\vec{B}_{i}$ must be perpendicular to $Δ \vec{s}_{i}$ and perpendicular to $\hat{r}$ . You should convince yourself that $\vec{B}_{i}$ in Figure 29.17 points in the correct direction. Notice that the y-component of $\vec{B}_{i}$ is canceled by the y-component of magnetic field $\vec{B}_{j}$ due to the current segment at the bottom of the loop, $180 °$ away. In fact, every current segment on the loop can be paired with a segment $180 °$ away, on the opposite side of the loop, such that the x- and y-components of $\vec{B}$ cancel and the components of $\vec{B}$ parallel to the z-axis add. In other words, the symmetry of the loop requires the on-axis magnetic field to point along the z-axis. Knowing that we need to sum only the z-components will simplify our calculation.

SOLVE We can use the Biot-Savart law to find the z-component ${(B_{i})}_{z} = B_{i} cos ϕ$ of the magnetic field of segment i. The cross product $Δ \vec{s}_{i} \times \hat{r}$ has magnitude $(Δ s) (1) sin 90 ° = Δ s$ , thus

{(B_{i})}_{z} = \frac{μ_{0}}{4 π} \frac{I Δ s}{r^{2}} cos ϕ = \frac{μ_{0} I cos ϕ}{4 π (z^{2} + R^{2})} Δ s

where we used $r = {(z^{2} + R^{2})}^{1 / 2}$ . You can see, because $ϕ + γ =$ $90 °$ , that angle $ϕ$ is also the angle between $\hat{r}$ and the radius of the loop. Hence $cos ϕ = R / r$ , and

{(B_{i})}_{z} = \frac{μ_{0} I R}{4 π {(z^{2} + R^{2})}^{3 / 2}} Δ s

The final step is to sum the magnetic fields due to all the segments:

B_{loop} = \sum_{i}^{} {(B_{i})}_{z} = \frac{μ_{0} I R}{4 π {(z^{2} + R^{2})}^{3 / 2}} \sum_{i}^{} Δ s

In this case, unlike the straight wire, none of the terms multiplying $Δ s$ depends on the position of segment i, so all these terms can be factored out of the summation. We’re left with a summation that adds up the lengths of all the small segments. But this is just the total length of the wire, which is the circumference $2 π R$ . Thus the on-axis magnetic field of a current loop is

B_{loop} = \frac{μ_{0} I R}{4 π {(z^{2} + R^{2})}^{3 / 2}} 2 π R = \frac{μ_{0}}{2} \frac{I R^{2}}{{(z^{2} + R^{2})}^{3 / 2}}

In practice, current often passes through a coil consisting of N turns of wire. If the turns are all very close together, so that the magnetic field of each is essentially the same, then the magnetic field of a coil is N times the magnetic field of a current loop. The magnetic field at the center $(z = 0)$ of an N-turn coil, or N-turn current loop, is

B_{coil center} = \frac{μ_{0}}{2} \frac{N I}{R} (N -turn current loop)

(29.8)

This is the second of our key magnetic field models.

EXAMPLE 29.6 Matching the earth’s magnetic field

What current is needed in a 5-turn, 10-cm-diameter coil to cancel the earth’s magnetic field at the center of the coil?

MODEL One way to create a zero-field region of space is to generate a magnetic field equal to the earth’s field but pointing in the opposite direction. The vector sum of the two fields is zero.

VISUALIZE FIGURE 29.18 shows a five-turn coil of wire. We will assume that the coil is thin and model it as a five-turn loop.

The figure illustrates the magnetic field at the center of a current-carrying coil.

FIGURE 29.18 Full Alternative Text

SOLVE The earth’s magnetic field, from Table 29.1, is $5 \times 10^{- 5} T$ . We can use Equation 29.8 to find that the current needed to generate a $5 \times 10^{- 5} T$ field is

I = \frac{2 R B}{μ_{0} N} = \frac{2 (0.050 m) (5.0 \times 10^{- 5} T)}{5 (1.26 \times 10^{- 6} T m / A)} = 0.80 A

REVIEW A 0.80 A current is easily produced. Although there are better ways to cancel the earth’s field than using a simple coil, this illustrates the idea.

29.4 The Magnetic Field of a Current

MODEL 29.1 Three key magnetic fields

FIGURE 29.13 Relating the charge velocity $\vec{v}$ to the current I.

PROBLEM-SOLVING STRATEGY 29.1
The magnetic field of a current

EXAMPLE 29.3 The magnetic field of a long, straight wire

FIGURE 29.14 Calculating the magnetic field of a long, straight wire carrying current I.

FIGURE 29.15 The magnetic field of a long, straight wire carrying current I.

EXAMPLE 29.4 The magnetic field strength near a heater wire

EXAMPLE 29.5 The magnetic field of a current loop

FIGURE 29.16 A current loop.

FIGURE 29.17 Calculating the magnetic field of a current loop.

EXAMPLE 29.6 Matching the earth’s magnetic field

FIGURE 29.18 A coil of wire.