Recent content by Silverado
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How Can I Use Magnetic Field Protractor to Solve for Magnetic Flux?
Thank you!- Silverado
- Post #7
- Forum: Introductory Physics Homework Help
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How Can I Use Magnetic Field Protractor to Solve for Magnetic Flux?
Take the derivative. So I can take the derivative of part (b). But theta needs to be expressed in terms of the time, t?- Silverado
- Post #5
- Forum: Introductory Physics Homework Help
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How Can I Use Magnetic Field Protractor to Solve for Magnetic Flux?
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)?- Silverado
- Post #3
- Forum: Introductory Physics Homework Help
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How Can I Use Magnetic Field Protractor to Solve for Magnetic Flux?
Homework Statement [/B] Homework Equations R = ρ (l/A) I = ε/R s = rθ Magnetic flux = Integral(B ⋅ dA) The Attempt at a Solution [/B] 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)...- Silverado
- Thread
- Field Magnetic Magnetic field
- Replies: 6
- Forum: Introductory Physics Homework Help
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Is My Method for Deflecting a Beam of Electrons Invalid?
Centripetal F = mv^2/r- Silverado
- Post #7
- Forum: Introductory Physics Homework Help
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Is My Method for Deflecting a Beam of Electrons Invalid?
The force is variable, therefore my method is invalid. How should I approach this problem, then?- Silverado
- Post #5
- Forum: Introductory Physics Homework Help
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Is My Method for Deflecting a Beam of Electrons Invalid?
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?- Silverado
- Post #3
- Forum: Introductory Physics Homework Help
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Is My Method for Deflecting a Beam of Electrons Invalid?
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 [/B] W = -U -Fd = U E1 = E2 K = U K = -Fd = -qvBsinθd K...- Silverado
- Thread
- Beam Electrons
- Replies: 7
- Forum: Introductory Physics Homework Help
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Power Dissipated in a Sliding Resistor
Ty for your help & I am glad to be a part of the community! :-)- Silverado
- Post #11
- Forum: Introductory Physics Homework Help
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Power Dissipated in a Sliding Resistor
Thanks again. Your help is very much appreciated.- Silverado
- Post #7
- Forum: Introductory Physics Homework Help
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Power Dissipated in a Sliding Resistor
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...- Silverado
- Post #3
- Forum: Introductory Physics Homework Help
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Power Dissipated in a Sliding Resistor
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...- Silverado
- Thread
- Power Power dissipated Resistor Sliding
- Replies: 10
- Forum: Introductory Physics Homework Help