Physics 21Fall, 2004Solution to HW-1328P 18Two long parallel wires 7.0 cm apart carry 16.5 Acurrents in the same direction. Determine the magnetic fieldstrength at a pointP12.0 cm from one wire and 13.0 cmfrom the other.αβααπ/2 − αBB1ββπ/2 − βB2The fieldBat the pointPwill be the vector sum of thefields from the two wires. We must find the components ofthe two fields so we can add them. We can use the Law ofCosines to solve for the anglesα:132= 122+ 72−2(12)(7) cosα⇒α= 81.79◦122= 132+ 72−2(13)(7) cosβ⇒β= 66.01◦The magnitude of the fields are obtained fromB=µ0I/2πR,whereR1= 0.012 m for the left wire andR2= 0.013 m forthe right wire. The diagrams show how we can use geometryto find the components ofB1andB2:B1=µ0I2π(0.12 m)−cos(π/2−α)ˆi+ sin(π/2−α)ˆjB2=µ0I2π(0.13 m)−cos(π/2−β)ˆi−sin(π/2−β)ˆjSubsituting the numbers, we findB=B1+B2= (−5.04ˆi−0.64ˆj)×10−5T28P 32A small loop of wire of radius 1.8 cm is placed at
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