# 5.2 The Biot–Savart Law

In the preceding section, we elected to use the magnetic flux density B to denote the presence of a magnetic field in a given region of space. We now work with the magnetic field intensity H instead. We do this in part to remind the reader that for most materials the flux and field are linearly related by B=μ H; therefore, knowledge of one implies knowledge of the other (assuming that μ is known).

Through his experiments on the deflection of compass needles by current-carrying wires, Hans Oersted established that currents induce magnetic fields that form closed loops around the wires (see Section 1-3.3). Building upon Oersted’s results, Jean Biot and Félix Savart arrived at an expression that relates the magnetic field H at any point in space to the current I that generates H. The Biot–Savart law states that the differential magnetic field d H generated by a steady current I flowing through a differential length vector d l is

d H=I 4 π d l×R^R 2 (A/m),(5.21)

where R=R^R is the distance vector between d l and the observation point P, as shown in Fig. 5-8. The SI unit for H is ampere⋅m/m 2=(A/m). It is important to remember that Eq. (5.21) assumes that d l is along the direction of the current I and the unit vector R^ points from the current element to the observation point.

# Figure 5-8 Magnetic field d H generated by a current element I d l. The direction of the field induced at point P is opposite to that induced at point P′.

Figure 5 8 Full Alternative Text

According to Eq. (5.21), d H varies as R−2, which is similar to the distance dependence of the electric field induced by an electric charge. However, unlike the electric field vector E, whose direction is along the distance vector R joining the charge to the observation point, the magnetic field H is orthogonal to the plane containing the direction of the current element d l and the distance vector R. At point P in Fig. 5-8, the direction of d H is out of the page, whereas at point P′ the direction of d H is into the page.

To determine the total magnetic field H due to a conductor of finite size, we need to sum up the contributions due to all the current elements making up the conductor. Hence, the Biot–Savart law becomes

where l is the line path along which I exists.

# 5.2.1 Magnetic Field Due to Surface and Volume Current Distributions

The Biot–Savart law also may be expressed in terms of distributed current sources (Fig. 5-9) such as the volume current density J, measured in (A/m 2), or the surface current density J s, measured in (A/m). The surface current density J s applies to currents that flow on the surface of a conductor in the form of a sheet of effectively zero thickness. When current sources are specified in terms of J s over a surface S or in terms of J over a volume υ, we can use the equivalence given by

I d l ⇔ J s d s ⇔ J d υ (5.23)

to express the Biot–Savart law as

# Figure 5-9 (a) The total current crossing the cross section S of the cylinder is I=∫S J⋅d s. (b) The total current flowing across the surface of the conductor is I=∫l J s d l.

(a) Volume current density J in A/m2

(b) Surface current density Js in A/m

# Example 5-2: Magnetic Field of a Linear Conductor

A free-standing linear conductor of length l carries a current I along the z axis as shown in Fig. 5-10. Determine the magnetic flux density B at a point P located at a distance r in the x−y plane. The wire is, of course, part of a closed-loop circuit, but our current interest is only in the conducting wire.

# Figure 5-10 Linear conductor of length l carrying a current I. (a) The field d H at point P due to incremental current element d l. (b) Limiting angles θ 1 and θ 2, each measured between vector I d l and the vector connecting the end of the conductor associated with that angle to point P (Example 5-2).

Figure 5 10-a Full Alternative Text
Figure 5 10-b Full Alternative Text

Solution: From Fig. 5-10, the direction of I is along +z^. Hence, the differential length vector is d l=z^d z, and d l×R^=d z (z^×R^)=ϕ^ sin θ^ d z, where ϕ^ is the azimuth direction and θ is the angle between d l and R^. Application of Eq. (5.22) gives

H=I 4 π∫z=−l/2 z=l/2 d l×R^R 2=ϕ^I 4 π∫−l/2 l/2 sin θ R 2 d z.(5.25)

Both R and θ are dependent on the integration variable z, but the radial distance r is not. For convenience, we convert the integration variable from z to θ by using the transformations

Upon inserting Eqs. (5.26a) and (5.26c) into Eq. (5.25), we have

H=ϕ^I 4 π∫θ 1 θ 2 sin θ r csc 2 θ d θ r 2 csc 2 θ=ϕ^I 4 π r∫θ 1 θ 2 sin θ d θ=ϕ^I 4 π r(cos θ 1−cos θ 2),(5.27)

where θ 1 and θ 2 are the limiting angles at z=−l/2 and z=l/2, respectively. From the right triangle in Fig. 5-10(b), it follows that

Hence,

B=μ 0 H=ϕ^ μ 0 I l 2 π r 4 r 2+l 2 (T).(5.29)

For an infinitely long wire with l≫r, Eq. (5.29) reduces to

# Figure 5-11 Magnetic field surrounding a long, linear current-carrying conductor.

# Example 5-3: Magnetic Field of a Circular Loop

A circular loop of radius a carries a steady current I. Determine the magnetic field H at a point on the axis of the loop.

Solution: Let us place the loop in the x−y plane (Fig. 5-12). Our task is to obtain an expression for H at point P(0,0,z).

We start by noting that any element d l on the circular loop is perpendicular to the distance vector R and that all elements around the loop are at the same distance R from P with R=a 2+z 2. From Eq. (5.21), the magnitude of d H due to current element d l is

d H=I 4 π R 2|d l×R^|=I d l 4 π(a 2+z 2),(5.31)

and the direction of d H is perpendicular to the plane containing R and d l. d H is in the r−z plane (Fig. 5-12); therefore, it has components d H r and d H z. If we consider element d l′ located diametrically opposite to d l, we observe that the z components of the magnetic fields due to d l and d l′ add because they are in the same direction, but their r components cancel because they are in opposite directions. Hence, the non-vanishing component of the magnetic field is along z only. That is,

d H=z^d H z=z^ d H cos θ=z^I cos θ 4 π(a 2+z 2)d l.(5.32)

For a fixed point P(0,0,z) on the axis of the loop, all quantities in Eq. (5.32) are constant, except for d l. Hence, integrating Eq. (5.32) over a circle of radius a gives

H=z^I cos θ 4 π(a 2+z 2)∮d l=z^I cos θ 4 π(a 2+z 2)(2 π a).(5.33)

# Figure 5-12 Circular loop carrying a current I (Example 5-3).

Figure 5 12 Full Alternative Text

Upon using the relation cos θ=a/(a 2+z 2)1/2, we obtain

At the center of the loop (z=0), Eq. (5.34) reduces to

H=z^I 2 a (at z=0),(5.35)

and at points very far away from the loop such that z 2≫a 2, Eq. (5.34) simplifies to

H=z^I a 2 2|z|3 (at|z|≫a).(5.36)

# 5.2.2 Magnetic Field of a Magnetic Dipole

In view of the definition given by Eq. (5.19) for the magnetic moment m of a current loop, a single-turn loop situated in the x−y plane (Fig. 5-12) has magnetic moment m=z^m with m=I π a 2. Consequently, Eq. (5.36) may be expressed as

H=z^m 2 π|z|3 (at|z|≫a).(5.37)

This expression applies to a point P far away from the loop and on its axis. Had we solved for H at any distant point P=(R,θ,ϕ) in a spherical coordinate system with R the distance between the center of the loop and point P, we would have obtained the expression

# Figure 5-13 Patterns of (a) the electric field of an electric dipole, (b) the magnetic field of a magnetic dipole, and (c) the magnetic field of a bar magnet. Far away from the sources, the field patterns are similar in all three cases.

(a) Electric dipole

Figure 5 13-a Full Alternative Text

(b) Magnetic dipole

Figure 5 13-b Full Alternative Text

(c) Bar magnet

Figure 5 13-c Full Alternative Text

Concept Question 5-5: Two infinitely long parallel wires carry currents of equal magnitude. What is the resultant magnetic field due to the two wires at a point midway between the wires—compared with the magnetic field due to one of them alone—if the currents are (a) in the same direction and (b) in opposite directions?

Concept Question 5-6: Devise a right-hand rule for the direction of the magnetic field due to a linear current-carrying conductor.

Concept Question 5-7: What is a magnetic dipole? Describe its magnetic field distribution.

# 5.2.3 Magnetic Force Between Two Parallel Conductors

In Section 5-1.1, we examined the magnetic force F m that acts on a current-carrying conductor when placed in an external magnetic field. The current in the conductor, however, also generates its own magnetic field. Hence, if two current-carrying conductors are placed in each other’s vicinity, each will exert a magnetic force on the other. Let us consider two very long (or effectively infinitely long), straight, free-standing, parallel wires separated by a distance d and carrying currents I 1 and I 2 in the z direction (Fig. 5-14) at y=−d/2 and y=d/2, respectively. We denote by B 1 the magnetic field due to current I 1, defined at the location of the wire carrying current I 2 and, conversely, by B 2 the field due to I 2 at the location of the wire carrying current I 1. From Eq. (5.30), with I=I 1, r=d, and ϕ^=−x^ at the location of I 2, the field B 1 is

B 1=−x^μ 0 I 1 2 π d.(5.39)

# Figure 5-14 Magnetic forces on parallel current-carrying conductors.

Figure 5 14 Full Alternative Text

The force F 2 exerted on a length l of wire I 2 due to its presence in field B 1 may be obtained by applying Eq. (5.12):

F 2=I 2 l z^×B 1=I 2 l z^×(−x^) μ 0 I 1 2 π d=−y^μ 0 I 1 I 2 l 2 π d,(5.40)

and the corresponding force per unit length is

F 2′=F 2 l=−y^μ 0 I 1 I 2 2 π d.(5.41)

A similar analysis performed for the force per unit length exerted on the wire carrying I 1 leads to

F 1′=y^μ 0 I 1 I 2 2 π d.(5.42)