Magnetic Field Inside a Charging Capacitor

  • Thread starter Plasmosis1
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Plasmosis1
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Homework Statement


A 10-cm-diameter parallel-plate capacitor has a 1.0 mm spacing. The electric field between the plates is increasing at the rate 1.4×106V/m*s.
What is the magnetic field strength 2.7cm from the axis?

Homework Equations



[itex]\oint[/itex]B*ds=[itex]\mu[/itex][itex]\epsilon[/itex]dE(t)/dt*A
B=([itex]\mu[/itex][itex]\epsilon[/itex]dE(t)/dt*A)/(2[itex]\pi[/itex]r)

B=([itex]\mu[/itex]Id)/(2[itex]\pi[/itex]R2)*r

The Attempt at a Solution


I tried using the first equation:
8.85E-12*4[itex]\pi[/itex]E-7*1.4E6*[itex]\pi[/itex](0.05)2/(2[itex]\pi[/itex]*0.027)=
7.2E-13T which is wrong
 

Answers and Replies

  • #2
36,100
13,024
"A" should be the area your integral encloses, not the total area of the capacitor.
 

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