chap32 - Chapter 32 Question 22 A coaxial cable consists of...

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Chapter 32 Question 22 A coaxial cable consists of an inner conductor of radius a and outer conductor of radius b, as shown in Fig. 32-21. Current flows along one conductor and back along the other. Show that the inductance per unit length of the cable is ) / ln( 2 0 a b π µ The coaxial cable can be viewed as a current loop. Current flows one way along the inner conductor, then at the end it connects to the outer conductor, then it flows all the way back to the other end, and then flows from the outer conductor back to the inner conductor. Since L is defined as I L Φ = we will need to find the flux through the current loop. The area of this loop is any cross section which the current flows around. Thus we can take a rectangle with length x, the length of the cable, and width b-a. Unfortunately the magnetic field is not uniform over this region, so we will have to calculate the flux with an integral. First we need the B field which can be found with Ampere’s law. For our Amperian loop we’ll use a circle of arbitrary radius r between a and b. That way it encloses the current of the inner wire, but not the current of the outer wire. r I B I r B I l d B π µ µ π µ 2 2 0 0 0 = = = v v This magnetic field curls around the wire, so it will slice through the area we are integrating over to produce the maximum flux. That means we can just take a scalar integral instead of a vector integral. For flux we’ll be integrating over BdA. We have already found B. For dA, we’ll use long rectangles with length x, the length of the coaxial cable, but very small width dr. Since the magnetic field only changes in the r direction, it will be constant all along x, and we only need to break it into tiny differential pieces in the r direction.
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