- #1
Simon777
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Homework Statement
A very long solenoid of circular cross section with radius a= 4.80 cm has n= 77.0 turns/cm of wire. An electron is sitting outside the solenoid, at a distance r= 5.30 cm from the solenoid axis. What is the magnitude of the force on the electron while the current in the solenoid is ramped up at a rate of 38.0 Amps/s?
Homework Equations
emf= -L * dI/dt
F=E*q
L for a solenoid=mu not*N^2*A/l
The Attempt at a Solution
From what another has told me, this is the sequence of steps that works, but I want to know why:
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The induced emf is μo*n*A*dI/dt = 4πx10^-7*7400*π*0.0540^2*36 = 3.07x10^-3V
So the electric field = V/2πr = 3.07x10^-3/(2π*0.0590) = 8.27x10^-3N/C
So the force on the electron = E*q = 8.27x10^-3N/C*1.60x10^-19C = 1.32x10^-21N
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I want to start from basics and build the above. So far I start with:
emf= -L * dI/dt
and inductance for a solenoid=mu not*N^2*A/l
so I make:
emf= -mu not*N^2*A/l * dI/dt
now N=number of turns and l=unit length for 1 turn so I can use the above info to turn N/l into the variable n to get rid of l.
So n=N/l and
emf= -mu not*N*(N/l)*A * dI/dt
now I can solve for emf, but I get stuck here because I don't know where E=V/(2pi r) comes from. I need E to plug into F=E*q to get the final answer. I know E=(q/A)/(2*epsilon not), but the only formula I know relating E to V is V= E*d, but I don't think it applies here.
Am I even on the right path to solving this?