Question

Problem regarding magnetic field strength, prompt posted below shortly.

Answer 1

A wire with mass per unit length 75 grams per meter runs horizontally at right angles to a horizontal magnetic field. A 6.2 Amp current in the wire results in its being suspended against gravity. What's the magnetic field strength?

Answer 2

The first thing I'd be thinking of doing would be finding the total mass of the wire based on its length so I can figure out F=mg, where mg is equal to the force suspending it against gravity, but I *can't* do that, because I'm not given an overall length, just a mass per unit length.

Answer 3

@Pompeii00

Answer 4

haha! they want you to be clever here!

Answer 5

Can you set up the entire equality for the magnetic force versus the weight of the wire in the most general way possible?

Answer 6

then maybe you'll see how you can get the answer

Answer 7

Alright, one sec, let me think about it.

Answer 8

\[qV\times B - ma=0\]

Answer 9

Okay, now try to write out the cross product as the magnitude of the cross product.

Answer 10

\[F_{B}-F_{g}=0\]

Answer 11

Yeah, if they give you the mass per unit length then you probably need to find something like the magentic field strength per unit length to counteract it. Or just go through it with a mass and hope that later it divides out.

Answer 12

Alright, will do.

Answer 13

\[qVB \sin(\theta)=qVB; \ \ \ qVB-mg=0.\]

Answer 14

Very nice!

Answer 15

Now move the mg to the right side, and then go ahead and rewrite the left in terms of different variables

Answer 16

Alright, and now just moving mg over to the other side and dividing by qV.

Answer 17

Yep, heh

Answer 18

\[B=\frac{mg}{qV}\]

Answer 19

okay... now qV can actually be rewritten as different quantities!

Answer 20

Their answer, though, is in the form \[\frac{IlB}{mg}\]

Answer 21

Which seems like a wholly different situation to me but I guess is the same thing/I dunno? Trying to figure that out

Answer 22

Whoops, reciprocal of what I just wrote, sorry

Answer 23

\[B=\frac{mg}{Il}\]Are these the equivalent substitutions?

Answer 24

Yes!

Answer 25

because you can rewrite

v = L/t

then move the t under the q, q/t = I

so qv = I L

Answer 26

\[V = IR; \ qV = qIR. \]
\[v=L/t, \]Lost you on that last formula there. Where did L/t come from?

Answer 27

velocity is length divided by time (meters/second)

Answer 28

Oh, I thought you meant Inductance for a second, capital L, and got all confused.

Answer 29

haha sorry, i was trying to distinguish... cuz I and I are the same, even though ones 'i' and one's L haha

Answer 30

but yeah, then you have m/L, which is your mass density

Answer 31

so you have all you need

Answer 32

Alright, yeah, this makes sense. Thank you. Needed to look back on the substitutions for a moment to make sure I got everything.