Posted by Spencer on Monday, April 7, 2008 at 4:51pm.
A 59.0 turn, 4.10cmdiameter coil with R = 0.530 {\Omega} surrounds a 1.70cmdiameter solenoid. The solenoid is 21.0 cm long and has 170 turns. The 60 Hz current through the solenoid is I_sol = (.470A)*sin(2*pi*f*t). I guess f is the frequency.
Find the value of I_coil, the induced current in the coil at time t= 1.85 seconds?
I tried plugging into solenoid equation getting field then using that field to find induced current but am not getting it.

physics EM  bobpursley, Monday, April 7, 2008 at 9:22pm
what did you plug into the solenoid equation? What did you take as dB/dt?
Let me see you work, I suspect the error is in the detail.

physics EM  Spencer, Monday, April 7, 2008 at 9:50pm
ok for dB/Dt I get .18025cos(2pift). for induced emf its Area*dB/dt*N(turns). ive tried it with a problem I know and it works fine. at the end do i just plug in the given t into the equation. Please could you work it out i think its calculation but i cant find it. I use B=(u*N*Isol)/L this gives magnetic field for solenoid. B= (4pie7*170*.470*sin(2pift)/.21
this gives 4.781e4*sin(2pift)
dB/dT is .18025 cos(2pift).
the area in the calculation is the area between the solenoid and coil. so its pi*((.041/20)^2(.017/2)^2)=area. so this area times the dB/dT times number of turns 59 should equal induced emf . then that divided by resistance should be it. could you work it out and tell me what you get. I keep getting 20.3 mA
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