Question

A 10 amp current passing through a wire .254m in length producing a side force of 5N
a) Find the magnitude of B require to produce the force
the closest equation I could find to solve part A is F=IL x B
it is alright to omit the vector and use standard operation to solve this?

Part B) If the wire is of copper with 0.003175 m in diameter, what would be the voltage drop and the power loss?

Answer 1

Yes, you can disregard the vector operation (i.e. the cross product) and treat it instead as a multiplication.

Answer 2

Thanks, that's what I been thinking but I haven't got around the question why. I suppose it's because it's in perpendicular?

As for Part B, if anyone have a hint please pitch in.

Answer 3

Yes it's because the B field is has to be perpendicular to the wire to produce a force that is also perpendicular to the wire. For part B, start with this:\[R=rho \frac{ L }{ A }\]where R is resistance; ρ is the electrical resistivity of copper; L is the wire length; and A is the cross sectional area of the wire.

Answer 4

Well, that didn't work. Here it is:\[R = rho \frac{ L }{ A }\]

Answer 5

That...made a lot of sense, thank you.

Answer 6

Last bump? for R, I got 5.4218*10^-4
Am a bit confused on the voltage drop and power loss.
I*R=V
Concept wise, could anyone explain what is happening in this problem?

Answer 7

You've got the right equation for voltage drop.

Power in an electric circuit is given by:\[P=VI\]where P is power; V is voltage; and I is current.

Answer 8

Note that that power equation has several forms:\[P=VI=\frac{ V ^{2} }{ R }=I ^{2}R\]

Answer 9

I feel like am asking way too much question for this one particular question. Thank you for answering all of these.
Just one last question, although I've been using the equation I*R=V a lot lately, modeling circuits and so forth, I can't seem to grasp the fundamental(?) of electromagnetic field and how voltage drop plays a role in this model.

Answer 10

For the sake of understanding the concept and appreciating all these formula (heheh) could you please explain.

Answer 11

Voltage drop is the same thing as voltage: it's the potential difference between two points. In this case, it's the potential difference between one side of the resistor and the other. In the case of a circuit with a battery, it's the potential difference between the battery's positive and negative terminals.

Voltage can be written as:\[V=Ed\]where V is voltage; E is the electric field strength; and d is the distance between two points in the electric field. If we use the formula for electric field we get this:\[V=Ed=\left[ \frac{ 1 }{ 4\pi \epsilon _{O} }\frac{ q _{1}q _{2} }{ r ^{2} } \right]r=\frac{ 1 }{ 4\pi \epsilon _{O} }\frac{ q _{1}q _{2} }{ r}\]

Answer 12

I think I see it now. Even though this deals with electromagnetic field(?) the property of voltage (by definition and practicality remain the same. THANKS!

Answer 13

For this problem, it's actually not necessary to know anything about the electric field.