# B w l 2 μ b sin90º w 2 μ b l 2 ia b l 2 il 2 b l 2

• doctorjune
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B!"WL2=μBsin90ºW=2μBL=2IA()BL=2IL2BL=2ILBW=2 3.6A()0.3m()0.4T()=0.864Nshared via CourseHero.comB=μoI2r±+RB!"=?R6This study resource wasshared via CourseHero.com
Now we need only to find the four currents. The current through segmentsBandCequal the current through thehorizontal resistor, and since the potential difference across that resistor is the voltage of the battery, we have simply:The current through conductorsAandDare the same as the current from the battery, and for this current we need tocompute the equivalent resistance.The resistors are in parallel, so the equivalent resistance is easy to find, and with itthe two currents:Putting it all together:BA=14μoIA2aout of page!()BB=14μoIB2aout of page!()BC=14μoIC2 2a()into page!()BD=14μoID2 2a()into page!()Btot=μo8aIA+IB12IC12IDout of page!()IB=IC=εR1Req=1R+1R=2RReq=R2IA=ID=εReq=2εRBtot=μo8a2εR+εR12εR122εRout of page!()=3μoε16aRout of page!()7This study resource wasshared via CourseHero.comPowered by TCPDF ()
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Term
Spring
Professor
PRESTI
Tags
Magnetic Field