- #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?