# Homework Help: Calculate the magnetic field at 2 points (checking my result)

1. Dec 13, 2009

### fluidistic

1. The problem statement, all variables and given/known data
See the picture for a clear description of the situation. I must find the magnetic field at point $$P_1$$ and $$P_2$$. The picture represent a transversal cut of a wire carrying a current I going off the page to our direction. The density of the current is J.

2. Relevant equations None given.

3. The attempt at a solution
I used Ampere's law. I know the figure doesn't seem enough symmetric to use Ampere's law, but I think it is. I'll explain why: I though the problem like if there were 2 wires. One of radius 2a and with a current I in our direction and another wire with radius a with current -I/4 (since the area of the "hole" of the wire is one fourth of the area of what would be a circular wire of radius 2a.)
Thanks to Ampere's law, $$B_1=\frac{\mu _0 I}{2 \pi d}$$.
$$B_1(P_1)=\frac{\mu _0 I}{2 \pi (2a+d)}$$ and $$B_1 (P_2)= \frac{\mu _0 I}{2 \pi (2a+d)}$$.

Now $$B_2 =-\frac{\mu _0 I}{8\pi d'}$$.
$$B_2(P_1)=-\frac{\mu _0 I}{8\pi (a+d)}$$ and $$B_2(P_2)=-\frac{\mu _0 I}{8 \pi (3a+d)}$$.
At last, $$B(P_1)=B_1(P_1)+B_2(P_1)$$ and $$B(P_2)=B_1(P_2)+B_2(P_2)$$.

Am I right?

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2. Dec 14, 2009

### kuruman

Your approach is correct, but your use of Ampere's Law is not. It is valid for points external to the wire. You need the field for a point inside the wire. If you draw an Amperian loop passing through that point, the current enclosed by the loop is less than the total current I, is it not? So what does Ampere's Law say in this case?

3. Dec 14, 2009

### fluidistic

You are absolutely right on the fact that my use of Ampere's law is valid only at exterior points of the wire. In fact $$P_1$$ and $$P_2$$ are exterior points.
The picture might be confusing now I realize. There's like a very small circle which says "I". In fact this circle only represent the sense of the current, the point in it means that it's going into our direction and from us to the sheet of paper.

Am I misunderstanding you?

4. Dec 14, 2009

### kuruman

Sorry, I missed that the two points are external.

You need to watch your signs. If B1 is positive at P1, then it must be negative at P2. At each point, B2 must have opposite sign to B1 because the current is in the opposite direction.

Last edited: Dec 14, 2009
5. Dec 14, 2009

### fluidistic

Yes I know, but why would it help me? I don't have to calculate the magnetic field at any point inside the wire.
Anyway, $$I=\pi (2a)^2 J \Rightarrow J=\frac{I}{4\pi a^2}$$ and because J is constant in all the wire, I', the current enclosed by any wire of radius r is worth $$I'=\frac{\pi r^2 I}{4\pi a^2}$$. But I don't see the application. I guess I'm still not understanding you. (I'm sorry about it!)

6. Dec 14, 2009

### kuruman

I edited my previous post. You did not misunderstand me, I misunderstood you.

7. Dec 14, 2009

### fluidistic

Ok no problem! I thank you for helping me!

My $$B_2$$ has opposed sign to $$B_1$$ in both $$P_1$$ and $$P_2$$. But I don't see why my $$B_1(P_1)$$ should be positive and not my $$B_1(P_2)$$.