Faraday's Law and electron force

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Ithryndil
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Homework Statement


For the situation described in Figure P31.32, the magnetic field changes with time according to the expression B = (8.00t^3 - 4.00t^2 + 0.800) T, and r2 = 2R = 5.00 cm.

Figure P31.32
(a) Calculate the magnitude and direction of the force exerted on an electron located at point P2 when t = 2.00 s.


Homework Equations


E = -0.5r(dB/dt)
F = qE + (qv X B)


The Attempt at a Solution


E = -0.5r(dB/dt)
F = qE + (qv X B)

Alright, the force equation should simplify to just F = qE because B is zero outside the circle.

When I plug in and get:
F = 0.5qr (24t^2-8t) and plug in for the values:
q = 1.602 x 10^(-19)
r = 0.05m
t = 2

I get: 3.204 x 10^(-19) which is wrong...

What am I doing wrong here?
 
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Hi Ithryndil,

Ithryndil said:

Homework Statement


For the situation described in Figure P31.32, the magnetic field changes with time according to the expression B = (8.00t^3 - 4.00t^2 + 0.800) T, and r2 = 2R = 5.00 cm.

Figure P31.32
(a) Calculate the magnitude and direction of the force exerted on an electron located at point P2 when t = 2.00 s.


Homework Equations


E = -0.5r(dB/dt)

I believe this is where your error is; this equation is not correct. How did you get it, and do you see what it needs to be?
 
The equation is right out of my book.
 
Ithryndil said:
The equation is right out of my book.

My guess is that in your book it is part of an example problem; in your diagram for example that equation would apply to the point P1, but not to point P2.

You need to derive an equation that applies to point P2, starting with Faraday's law.
 
Alright, yeah looks the equation applies to inside the circle, not to the outside...alright, well it's late, I will work on that derivation tomorrow after some sleep. Thanks for the insight.
 
Thank you. I got it. The expression of E I needed is:

E = [R^2(24.0t^2 - 8.00t)]/(2r) where R is the radius of the circle and r is the radius of the electron's orbit.