9-4 02wire wireˆ4Iddrµπ×==∫∫srBB (9.1.4) The integral is a vector integral, which means that the expression for Bis really three integrals, one for each component of B. The vector nature of this integral appears in the cross productˆId ×sr. Understanding how to evaluate this cross product and then perform the integral will be the key to learning how to use the Biot-Savart law. Interactive Simulation 9.1: Magnetic Field of a Current Element Figure 9.1.2 is an interactive ShockWave display that shows the magnetic field of a current element from Eq. (9.1.1). This interactive display allows you to move the position of the observer about the source current element to see how moving that position changes the value of the magnetic field at the position of the observer. Figure 9.1.2 Magnetic field of a current element. Example 9.1: Magnetic Field due to a Finite Straight Wire A thin, straight wire carrying a current I is placed along the x-axis, as shown in Figure 9.1.3. Evaluate the magnetic field at point P. Note that we have assumed that the leads to the ends of the wire make canceling contributions to the net magnetic field at the point P . Figure 9.1.3 A thin straight wire carrying a current I.9-5Solution: This is a typical example involving the use of the Biot-Savart law. We solve the problem using the methodology summarized in Section 9.10. (1) Source point (coordinates denoted with a prime) Consider a differential element ˆ'ddx=+si carrying current I in the x-direction. The location of this source is represented by ˆ''x=ri. (2) Field point (coordinates denoted with a subscript “P”) Since the field point P is located at (, ) (0,)xya=, the position vector describing P is ˆPa=rj. (3) Relative position vector The vector 'P=−rr r is a “relative” position vector which points from the source point to the field point. In this case, ˆˆ'ax=−rji, and the magnitude22|| 'rax== +ris the distance from between the source and P. The corresponding unit vector is given by 22ˆˆ'ˆˆˆsin cos'axraxθθ−== = −+rjirji (4) The cross product ˆd ×sr The cross product is given by ˆˆˆˆˆ( ' ) ( cos sin ) ( 'sin )ddx dxθθθ×=+ ×− + =sr i ijk (5) Write down the contribution to the magnetic field due to Id s The expression is 0022ˆsinˆ44IIddxdrrµµθππ′×==srBk which shows that the magnetic field at P will point in the ˆ+k direction, or out of the page. (6) Simplify and carry out the integration9-6The variables θ, x’ and r are not independent of each other. In order to complete the integration, let us rewrite the variables x’ and r in terms of θ. From Figure 9.1.3, we have 2/sin( ) csccot( ) cot cscra axaadxadπθ θπθθ θθ=−=⎧⎪⎨′′=−=− ⇒=⎪⎩ Upon substituting the above expressions, the differential contribution to the magnetic field is obtained as 2002(csc )sinˆˆsin4(csc) 4IIadddaaµµθθ θθθπθ π==Bkk Integrating over all angles subtended from 1θ to 2πθ− (note our definition of 2θ), we obtain []210021021ˆˆsin cos( ) cos44ˆ(cos cos )4IIdaaIaπθθµµθθ π θ θππµθθπ−==−−−=+∫Bk kk (9.1.5) The first term involving 2θ accounts for the contribution from the portion along the +x axis, while the second term involving 1θcontains the contribution from the portion along the x− axis. The two terms add! Let’s examine the following cases: (i) In the symmetric case where 21θθ=, the field point P is located along the perpendicular bisector. If the length of the rod is 2L , then 221cos /LLaθ=+ and the magnetic field is 00122cos22IILBaaLaµµθππ==+ (9.1.6) (ii) The infinite length limit L →∞ This limit is obtained by choosing 12(, ) (0,0)θθ=. The magnetic field at a distance a away becomes 02IBaµπ=
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