ps12sol

ps12sol - 8.02 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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PS12-1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 12 Solutions Analytic Problems… Problem 1: Coaxial Cable and Power Flow A coaxial cable consists of two concentric long hollow cylinders of zero resistance; the inner has radius a , the outer has radius b , and the length of both is l , with lb >> , as shown in the figure. The cable transmits DC power from a battery to a load. The battery provides an electromotive force ε between the two conductors at one end of the cable, and the load is a resistance R connected between the two conductors at the other end of the cable. A current I flows down the inner conductor and back up the outer one. The battery charges the inner conductor to a charge Q and the outer conductor to a charge Q + . (a) Find the direction and magnitude of the electric field E G everywhere. Consider a Gaussian surface in the form of a cylinder with radius r and length l , coaxial with the cylinders. Inside the inner cylinder ( r < a ) and outside the outer cylinder ( r > b ) no charge is enclosed and hence the field is 0. In between the two cylinders ( a < r < b ) the charge enclosed by the Gaussian surface is – Q , the total flux through the Gaussian cylinder is (2 ) E dE r l π Φ= ⋅ = ∫∫ EA G G w Thus, Gauss’s law leads to enc 0 ) q Er l = , or enc 0 ˆˆ (inward) for , 0 elsewhere 22 q Q arb rl ππ == < < r G (b) Find the direction and magnitude of the magnetic field B G everywhere. Just as with the E field, the enclosed current I enc in the Ampere’s loop with radius r is zero inside the inner cylinder ( r < a ) and outside the outer cylinder ( r > b ) and hence the field there is 0. In between the two cylinders ( a < r < b ) the current enclosed is – I . Applying Ampere’s law, 0e n c ) dB r I πµ ⋅= = Bs G G v , we obtain 0 ˆ (clockwise viewing from the left side) for , 0 elsewhere 2 I r µ =− < < B G ϕ
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PS12-2 (c) Calculate the Poynting vector S G in the cable. For arb << , the Poynting vector is 0 22 00 0 0 11 ˆ ˆˆ 4 I QI Q rl r µ µµ π ε πε ⎛⎞ = × = ⎜⎟ ⎝⎠ SE B r k G GG ϕ (from right to left) On the other hand, for ra < and rb > , we have 0 = S G . (d) By integrating S G over appropriate surface, find the power that flows into the coaxial cable. With () ˆ 2 dr d r = Ak G , the power is 0 0 1 (2 ) ln 2 4 b a S b Pd r d r la lr =⋅ = = ∫∫ SA w (e) How does your result in (d) compare to the power dissipated in the resistor? Since ln b a QQ b dd r I R l a =⋅= = = Es G G the charge Q is related to the resistance R by 0 2 ln( / ) lIR Q ba = . The above expression for P becomes 2 0 0 2 ln ln( / ) 2 lIR I b PI R l a == which is equal to the rate of energy dissipation in a resistor with resistance R . Problem 2: Electromagnetic Plane Wave An electromagnetic plane wave is propagating in vacuum has a magnetic field given by 0 10 1 ˆ ( ) 0 u B f ax bt f u otherwise < < =+ = Bj G where a and b are positive quantities. The ˆ + k direction is out of the paper.
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ps12sol - 8.02 MASSACHUSETTS INSTITUTE OF TECHNOLOGY...

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