Yes, You're right. That law goes as follows:
ε = - the rate of change of flux (sorry I do not know how to write this in the proper notation)
And the induced current is found, simply, through Ohm's law: ε = iR
So how can I, with this, go about part (c)?
Homework Statement
[/B]
Homework Equations
R = ρ (l/A)
I = ε/R
s = rθ
Magnetic flux = Integral(B ⋅ dA)
The Attempt at a Solution
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I managed to solve parts (a) and (b) independently. Here is my work for those two parts:
My real issue is with part (c) which uses the answers from (a)...
The electron moves up and to the left in a parabolic trajectory. Therefore, there must be a force exerted up and a force exerted left which negates the electrons initial kinetic energy. How does this realization affect my derivation?
Homework Statement
Homework Equations
W = -U for conservative forces
F = qV x B = qvBsinθ
W = Fd for conservative forces
K = 1/2mv^2
sinθ = 1 for 90° angles
E1 = E2 (conservation of mechanical energy)
The Attempt at a Solution
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W = -U
-Fd = U
E1 = E2
K = U
K = -Fd = -qvBsinθd
K...
I tried the problem again using your advice and I think I am doing too much algebra. I used your info. in addition to the loop and junction rule. I reconstructed the circuit with two loops, one containing resistors R and Ra and the other Ra and Rb.
JR:
I1 = I2 + I3
LR:
Upper loop: -RI3 + RaI2...
Homework Statement
The circuit [http://tinypic.com/r/2cwl7bt/9] shows a battery connected across a uniform resistor R0. A sliding contact can move across the resistor from x = 0 at the left to x = 10 cm at the right. Find an expression for the power dissipated in the resistor R as a function...