Magnetic Field

[mspeedwagon]

Last one for the day...

Two long parallel wires carry currents of 20A and 5A in opposite directions. The wires are separated by 0.2 m. What is the magnetic field midway between the two wires?

The answer is either 5.0 x 10^5 or 10^-5... need to get new ink... negative signs didn't print out.

I thought this was a simple two step problem. First solve for F/l and then solve for B... but I'm missing something. Any takers. Thanks in advance.

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[Jepstein30]

Last one for the day...

Two long parallel wires carry currents of 20A and 5A in opposite directions. The wires are separated by 0.2 m. What is the magnetic field midway between the two wires?

The answer is either 5.0 x 10^5 or 10^-5... need to get new ink... negative signs didn't print out.

I thought this was a simple two step problem. First solve for F/l and then solve for B... but I'm missing something. Any takers. Thanks in advance.

Haven't really seen this type of problem on any FLs so I think it is a little out of the scope of the MCAT (i.e. they'll give you an equation if its on there) but I do remember a little of this from physics.

B = uI/2(pi)r where u is the permittivity of free space and r the distance between the two wires.

Not sure how you got that answer though because you'd have to plug in 50 for I and don't see why you would do that... I'm pretty sure that IS the equation though! Obviously forgot how to use it haha

 

[mspeedwagon]

Figured this may be just out of scope. Supplementing EK physics with a few other books. Have a long way to go to my target score in Jan... good thing is I've got a few months.

 

[rjosh33]

Figured this may be just out of scope. Supplementing EK physics with a few other books. Have a long way to go to my target score in Jan... good thing is I've got a few months.

Yeah, methinks this would definitely be out of the scope of the MCAT, but it is solvable using the formula Jepstein mentioned. The key is that there are two fields created (one from each wire), and each field must be calculated separately and then added together.

So, from a long current-carrying wire, the magnetic field is B = [(mu)(I)/2(pi)r], where mu is 4(pi) x 10^-7, I is the current, and r is the distance in space from the wire (in meters of course).

For the 20 A wire:

B = 80(pi)x10^-7/(2(pi)(0.1))

For the 5 A wire:

B = 20(pi)x10^-7/(2(pi)(0.1))

Now that we have the magnitudes, we need to decide if these magnetic fields are additive or destructive. Using the right hand rule for each wire, we can see that the magnetic field for each one points in the same direction at the point midway between them. Because each field is in the same direction, we add the two magnetic fields together.

So, simplifying the denominators a little and then adding gives us a total magnetic field at the midway point of:

B = 100(pi)x10^-7/(0.2(pi))

Pi will cancel, and 100 divided by 0.2 is 500. So, we now have

B = 500 x 10^-7. This is equivalent to 5x10^-5 T. There's your final answer.

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ETA: My guess is that if current-carrying wires came up on the MCAT, it would be a qualitative question. Something along the lines of:

"A long wire that carryies current of 10 A is placed next to a wire carrying 15 A in the same direction. The 10 A wire will be:

A. Repelled from the 15 A wire.
B. Attracted to the 15 A wire.
C. Rotated 90 degrees counterclockwise.
D. Rotated 90 degrees clockwise."