Question

A 59.0 -turn, 4.10-cm-diameter coil with R = 0.530 {\Omega} surrounds a 1.70-cm-diameter 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.

Answers

bobpursley
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.
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= (4pie-7*170*.470*sin(2pift)/.21
this gives 4.781e-4*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
please help by 9am.
nvm found it elsewhere. turns out i had my calc in degrees instead of radians. don't you just hate when that happens.

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