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SECTION 3 0.1 • The Biot–Savart Law 929 Figure 30.3 (Example 30.1) (a) A thin, straight wire carrying a current I. The magnetic field at point P due to the current in each element ds of the wire is out of the page, so the net field at point P is also out of the page. (b) The angles ! 1 and ! 2 used for determining the net field. When the wire is infinitely long, ! 1 " 0 and ! 2 " 180°. (a) O x ds I θ rˆ r a P ds = dx x (b) θ 1 P θ 2 θ θ 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 current Solution From the Biot–Savart law, we expect that the s ! ˆr is out of the page. In fact, because all of the current elements I d s lie in the plane of the page, they all produce a magnetic field kˆ being a unit vector pointing out of the page, we see that where represents the magnitude of d s ! rˆ . Because rˆ is a unit vector, the magnitude of the cross "ds ! ˆr" d s ! ˆr " "ds ! ˆr" ˆk " (dx sin !) ˆk product is simply the magnitude of d s, which is the length dx. Substitution into Equation 30.1 gives Because all current elements produce a magnetic field kˆ direction, let us restrict our attention to the magnitude of the field due to one current element, which is To integrate this expression, we must relate the variables !, Because tan ! " a/(' x) from the right triangle in Figure s is located at a negative value of x), we have x " 'a cot ! Taking the derivative of this expression gives (3) dx " a csc 2 ! d! Substitution of Equations (2) and (3) into Equation (1) gives an expression in which the only variable is !. We now obtain 1 to ! 2 as defined in Figure 30.3b: (30.4) We can use this result to find the magnetic field of any straight current-carrying wire if we know the geometry and 1 and ! 2 . Consider the special case of an infinitely long, straight wire. If we let the wire in Figure 1 " 0 and ! 2 " $ for length elements ranging between positions x " ' ) and 1 ' cos ! 2 ) " (cos 0 ' cos $) " 2, Equation 30.4 becomes (30.5) Equations 30.4 and 30.5 both show that the magnitude of # 0 I 2$a B " # 0 I 4$a (cos ! 1 ' cos ! 2 ) B " # 0 I 4$a
! ! 2 ! 1 sin ! d! " (4)
dB " # 0 I 4$
a csc 2 ! sin ! d! a
2 csc 2 ! " # 0 I 4$a sin ! d! (2)
r " a sin ! " a csc ! (1)
dB " # 0
I 4$
dx sin ! r
2 d
B " (dB)ˆk " # 0 I 4$
dx sin ! r
2 ˆ k Interactive |