S ECT I O N 3 0 . 1 • The Biot–Savart Law 929

Figure 30.3 (Example 30.1) (a) A thin, straight wire carrying acurrent I. The magnetic field at point P due to the current ineach element ds of the wire is out of the page, so the net field atpoint P is also out of the page. (b) The angles !1 and !2 usedfor determining the net field. When the wire is infinitely long,!1 " 0 and !2 " 180°.

(a)

Ox

dsI

θr̂

r a

Pds = dx

x

(b)

θ1

P

θ2θθ

y

At the Interactive Worked Example link at http://www.pse6.com, you can explore the field for different lengths of wire.

Example 30.1 Magnetic Field Surrounding a Thin, Straight Conductor

Consider a thin, straight wire carrying a constant currentI and placed along the x axis as shown in Figure 30.3.Determine the magnitude and direction of the magneticfield at point P due to this current.

Solution From the Biot–Savart law, we expect that themagnitude of the field is proportional to the current inthe wire and decreases as the distance a from the wire topoint P increases. We start by considering a lengthelement ds located a distance r from P. The direction ofthe magnetic field at point P due to the current in thiselement is out of the page because ds ! r̂ is out of thepage. In fact, because all of the current elements I ds liein the plane of the page, they all produce a magnetic fielddirected out of the page at point P. Thus, we have thedirection of the magnetic field at point P, and we needonly find the magnitude. Taking the origin at O andletting point P be along the positive y axis, with k̂ being aunit vector pointing out of the page, we see that

where represents the magnitude of ds ! r̂ .Because r̂ is a unit vector, the magnitude of the cross

"ds ! r̂ "

ds ! r̂ " "ds ! r̂ " k̂ " (dx sin !)k̂

product is simply the magnitude of ds, which is the lengthdx. Substitution into Equation 30.1 gives

Because all current elements produce a magnetic fieldin the k̂ direction, let us restrict our attention to themagnitude of the field due to one current element, which is

To integrate this expression, we must relate the variables !,x, and r. One approach is to express x and r in terms of !.From the geometry in Figure 30.3a, we have

Because tan ! " a/(' x) from the right triangle in Figure30.3a (the negative sign is necessary because ds is located ata negative value of x), we have

x " 'a cot !

Taking the derivative of this expression gives

(3) dx " a csc2 ! d!

Substitution of Equations (2) and (3) into Equation (1) gives

an expression in which the only variable is !. We now obtainthe magnitude of the magnetic field at point P by integrat-ing Equation (4) over all elements, where the subtendingangles range from !1 to !2 as defined in Figure 30.3b:

(30.4)

We can use this result to find the magnetic field of anystraight current-carrying wire if we know the geometry andhence the angles !1 and !2. Consider the special case of aninfinitely long, straight wire. If we let the wire in Figure30.3b become infinitely long, we see that !1 " 0 and !2 " $for length elements ranging between positions x " ' ) andx " * ). Because (cos !1 ' cos !2) " (cos 0 ' cos $) " 2,Equation 30.4 becomes

(30.5)

Equations 30.4 and 30.5 both show that the magnitude ofthe magnetic field is proportional to the current anddecreases with increasing distance from the wire, as weexpected. Notice that Equation 30.5 has the same mathe-matical form as the expression for the magnitude of theelectric field due to a long charged wire (see Eq. 24.7).

#0I2$a

B "

#0I4$a

(cos !1 ' cos !2)B "#0I4$a

!!2!1

sin ! d! "

(4) dB " #0I4$

a csc2 ! sin ! d!

a 2 csc2 !"

#0I4$a

sin ! d!

(2) r " asin !

" a csc !

(1) dB " #0 I4$

dx sin !

r 2

d B " (dB)k̂ "#0I4$

dx sin !

r 2 k̂

Interactive