# Calculating a magnetic field at a given point

1. Oct 31, 2009

### fluidistic

1. The problem statement, all variables and given/known data

For a picture of the problem, look at the attached picture.
There's a current i through the wire.
1)Calculate the magnetic field (at point C) produced by each segment of length L.
2)The semi-circular segment of radius R.
3)The entire wire.
2. Relevant equations
None given. Biot-Savart and Ampère's law.

3. The attempt at a solution
1)Using Biot-Savart law, I have that $$\vec B = \frac{\mu _0 i}{4 \pi} \int d \vec l \times \frac{\vec r}{r^3}$$ but $$\vec l$$ and $$\vec r$$ are parallel, so the cross product is worth $$0$$ and so the contribution of the 2 segments of length $$L$$ to the magnetic field at point $$C$$ is null.
2) and 3) : Using Ampère's law I get that $$B=\frac{\mu _0 i}{\pi R}$$

Is that right? I find my answer to 1) strange.

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2. Oct 31, 2009

### ideasrule

Your answer to 1) is right: charges moving directly towards or away from a point don't induce any magnetic field at that point. Your answer to 2) is a bit strange, though. How did you use Ampere's law? Which closed loop did you integrate around? I find it easier to just use the Biot-Savart law; the integral isn't necessary because the vector dl is always perpendicular to the vector r.

3. Oct 31, 2009

### fluidistic

Thank you very much for helping!
Ok, nice to know I got part 1) right.
2) $$\oint \vec B d \vec l = \mu _0i$$ but as you pointed out $$\vec B \perp \vec l$$ for the entire semi circle loop so that $$B \cdot \pi R = \mu _0 i$$, hence my result.
I can say I've took a surrounding to the semi circle loop, closed loop. Does it matter? The current enclosed will always be worth $$i$$.

4. Oct 31, 2009

### ideasrule

There are a few problems with this. First, there's no obvious symmetry that suggests B is the same along the entire loop. Second, even if B doesn't vary, you're trying to find the magnetic field at point C; finding the field along the loop doesn't help. Third, i represents the current enclosed within the loop; in your scenario, the current is flowing parallel to the loop you chose. Finally, you chose a semi-circle, not a closed loop. Ampere's law only works for closed loops (notice the circle on the integral sign in your equation). There are absolutely no exceptions.

5. Nov 1, 2009

### fluidistic

First, thanks for helping.

hmm I don't understand why not. The semi-circular loop is the only part of the wire that create the magnetic field at point C, so enclosing it by a loop sounded good to me.

I haven't been clear enough. :shy: sorry. I meant I can say I have chosen a closed loop which has the same shape as the semi-circular part of the wire. For example a loop that has a width of $$\varepsilon$$ arbitrary small or big, instead of no width as is considered the wire. But a closed loop that enclose absolutely all the semi-circular wire.

Lastly, I will see a friend today, I will ask him if he knows how to solve this problem. I will post back here anyway, whether or not I have to procedure to solve the problem. Answer to my misconception if you like though, I'd be glad to learn.

6. Nov 1, 2009

### Staff: Mentor

As ideasrule states, Ampere's law is not useful in solving this problem. Instead, use the Biot-Savart law for all parts of the problem. The integration is trivial.

7. Nov 1, 2009

### fluidistic

You're right, I've underestimated this part.
$$\vec B = \frac{\mu _0 i}{4 \pi} \oint d\vec l \times \frac{d \vec r}{r^3}$$ but as $$\vec l \perp \vec r$$, the path integral is worth $$\frac{\pi R}{R^2}=\frac{\pi}{R}$$, thus $$B= \frac{\mu _0 i}{4 R}$$.
The direction of the magnetic field must point upward C, I believe. (for symmetrical reasons).

I hope I got it right.

8. Nov 1, 2009

### Staff: Mentor

Looks good. (The direction of the field depends on the direction of the current. It's either into or out of the paper.)

9. Nov 1, 2009

### fluidistic

Ah I see.

Thanks a lot to both!