ch28-p048

# ch28-p048 - 48. We establish coordinates such that the two...

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axis. We take G B to be in the same direction as that of the current flow in the hypotenuse. Then, with BB == G 0 0750 .T , cos 0.0692T , sin 0.0288T. xy B B θ =− = = (a) Eq. 28-26 produces zero force when G G LB || so there is no force exerted on the hypotenuse of length 130 cm. (b) On the 50 cm side, the B x component produces a force iB yx A ± k, and there is no contribution from the B y component. Using SI units, the magnitude of the force on the A y side is therefore 4 00 0500 0 0692 0138 .. . . Am T N . bg b g b g = (c) On the 120 cm side, the B y component produces a force A ± k, and there is no contribution from the B x component. The magnitude of the force on the A x side is also 4 00 120 0 0288 . . T N . b g = (d) The net force is iB iB AA ±± , kk += 0 keeping in mind that B x < 0 due to our initial assumptions. If we had instead assumed G B went the opposite direction of the current flow in the hypotenuse, then
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## This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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