Recent content by Ozmahn

That makes a lot of sense, thank you both!

Homework Statement Six identical light bulbs (A-F) are connected in the following circuit. The resistance of each bulb is R and the idealistic battery has an emf of E. The switch K remains open (not connected) for questions (a) through (c)...

Okay awesome, that makes sense to me now. Thank you!

I hadn't even considered the normal force...looking at the force diagram, the normal force would cancel the force due to gravity, leaving the net force as only the magnetic force. Is that accurate?

So if i disregard the contribution of gravity to the net force, i just get 4.8/.027=17.7, much closer to the answer provided by masteringphyics. But i don't understand why you wouldn't consider gravity...is there an underlying assumption that i missed?

Sorry, I didn't clarify. They want to know the acceleration right after the switch is flipped. I thought that the emf is only induced as a result of changing current...if we're just looking at the instant that the switch is flipped, then there wouldn't be an induced emf yet since the bar hasn't...

Homework Statement A 2.60-N metal bar, 1.50 m long and having a resistance of 10.0 Ω, rests horizontally on conducting wires connecting it to the circuit shown in the figure(Figure 1) . The bar is in a uniform, horizontal, 1.60-T magnetic field and is not attached to the wires in the circuit...

Ohh that angle is 60°, and cos(60)=sin(30). I just need to make sure I'm using the correct angle...thanks for pointing that out!

The area vector is the vector that is perpendicular to the area, so the green vector is ##\vec {dA}## The rest of the equation makes sense, I think. I'm just having trouble putting together the electric field portion. A is the area of the face ##\epsilon_0## is constant E should be net field...

Okay this is my interpretation...the angle made between ## \vec {E_1} ## and ## \vec {dA} ## is 30°, as given by the problem. Now that I made this picture, I think I can see it. But just to make sure, sin(30) here is ##{\frac{E_{1y}}{E_{1x}}}## So from there, I shouldn't have to take into...

Homework Statement The electric field ##\vec E_1 ## one face of a parallelepiped is uniform over the entire face and is directed out of the face. At the opposite face, the electric field ##\vec E_2 ## is also uniform over the entire face and is directed into that face. The two faces in question...

I must have written that on my paper by accident, it should actually be 0.0475. I used the number because that's how my professor did the problem, but if I'm understanding what you're saying, the position of the object is x=0. And since it's centered directly above the center of the wire, 0.095...

I put the .06 back in there and integrated over the bounds -.0495 and .0495, but I'm pretty sure those bounds are wrong too because I get 4671.25N/C. The bounds should be the length of the wire, shouldn't they?

I'm having some real issues with this one.. I started over again, so here is what I've got so far: I know the y components cancel, so I'm left looking for the sum of all the x components. I'm switching from dy to dx to make things easier for myself, since the distance along the wire can also be...

Oh we did do an example of that in class. Okay so starting over... The wire will be broken up into infinitely small segments called dL. dQ will be ##\lambda dy##, and r will just be the distance from the center of the wire to point P, which is 0.06 away. I didn't understand this next part very...