A few physics questions?
@theEric
I can take a look infrequently! I'd like to help more, but I have lots of studying to do!
Okay! :)
A ferromagnetic material is subjected to strong magnetic field and afterward acts as permanent magnet. Why?? Long-range electric currents circulating iron Motion of iron nuclei makes magnetic field Magnetic field aligns magnetic domains mostly in a single direction Magnetic field gives ferromagnetic material electric charge I think its C?
Yep!
cool :) Is it false that magnetic field lines are perpendicular to direction of magnetic field at every point? I think it is false
It is false!
nice :) What shape do magnetic field lines have around a wire with a steady current in it? Lines parallel to direction of wire spirals converging inward toward wire concentric circles around wire none of the above I think its B?
Nope! They do not converge inward, that I've learned. And you don't see them do so, so they can't be spiraling, either. Do you have another guess?
straight lines radiating outward from wire is actually another option I almost left it out
Nope! Did you check out the link I posted? It's an image search!
oh nvm I think its conecentric circles around the wire :)
Yup! :)
First you need to find the formula for the magnetic field generated by a solenoid! I have to go! I'll be back soon. http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/solenoid.html You double the current. So, what happens to the magnetic field?
Did you find your answer?
You're on the right track! \(B=\mu nI\) Next, \(I\) is doubled, and \(\mu\) and \(n\) stay the same. We'll let this new \(I\) be \(I_2\). And let \(B_2\) be the next magnitude of the magnetic field. So, \(B_2=\mu nI_2\) \(I_2\) doubled \(I_1\), so \(I_2=2I_1\). So, really, \(B_2=\mu n 2I_1\) So, \(B_2=2(\mu n I_1)=2B_1\)
These kind of problems come up every once in a while. Like, electrostatic force - what happens when you double the radius? \(F_1=k\dfrac{q_1\ q_2}{r_1^2}\) so \(F_2=k\dfrac{q_1\ q_2}{r_2^2}=k\dfrac{q_1\ q_2}{(2r_1)^2}\\=k\dfrac{q_1\ q_2}{2^2r_1^2}=\dfrac14\left(k\dfrac{q_1\ q_2}{r_1^2}\right)=\dfrac14F_1\) General method: Make the substitution of the change you know, and then get the expression to look how it did before.
Well, I guess I didn't finish the last one. \(B_2=2B_1\), though, so you can substitute your value for \(B_1\) and calculate \(B_2\). As for the maglev, the wording is weird, but it's definitely not the last two! So I agree!
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