Physics 21Fall, 2004Solution to HW-1529P 21Determine the magnetic ﬂux through a square loopof sideaif one side is parallel to, and a distanceafrom, astraight wire that carries a currentI.The field of the wire isB(r) =µ0I/2πr, directed perpen-dicular to the loop. The ﬂux isΦ =2aaB(r)a dr=µ0I2πa2aadrrThe definite integral is2aadrr= lnr|2aa= ln 2a−lna= ln2aa= ln 2Combining terms, we obtainΦ =µ0Ia2πln 2≈0.6931µ0Ia2π.One can get a pretty good approximate answer by assumingB(r) is constant over the loop and taking the value at themiddle, wherer= 1.5a. ThenΦ =B(1.5a)a2=µ0I2π(1.5a)a2=23µ0Ia2π≈0.6667µ0Ia2π.29P 40Neon signs require 12 kV for their operation. Tooperate from a 220 V line, what must be the ratio of sec-ondary to primary turns of the transformer? What would bethe voltage output if the transformer were connected back-ward?Given thatVP= 12 kV andVS= 220 V, we can useVSVP=12,000220=NSNP= 54.5
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