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In all the problems, the wires carry current.

Problem A (2 wires going into the page, find net mag. field at point P )

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Problem B (A loop of wire, find net mag. field at point C)

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Problem C (2 Wires with current opposite directions, find mag. field at point C)

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My impression is that Bio-Savart is only used for finding the magnetic field of a wire at a point, where as the equation $B=\left(\frac{\mu_0}{2\pi}\right)\frac{I}{r}$ is used to find the magnetic field of a wire on another wire.

So in problem A, you can think of point P as another wire going into the page or out (doesn't matter). So you use $B=\left(\frac{\mu_0}{2\pi}\right)\frac{I}{r}$.

In problem B, you are finding the net magnetic field of 2 wires (the other 2 sides of the loop can be ignored) on a point. So you use Biot-Savart. With Biot-Savart you get $B_{net}$=$\frac{\mu_0I\theta}{4\pi}\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$. With $B=\left(\frac{\mu_0}{2\pi}\right)\frac{I}{r}$, you get a different result.

But the real difficulty is in deciding which to use for problem C, the question asks for the net magnetic field of 2 wires at a field between them. In this case, unlike in problem A, you can't think of the point as a wire (in or out of the page). My professor uses $B=\left(\frac{\mu_0}{2\pi}\right)\frac{I}{r}$.
Doesn't the assumption that the wires are of infinite length affect the calculation?

Edit: Here is my professor's answer key to problem C. enter image description here

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  • $\begingroup$ That's the prof's answer to problem B, is it not? $\endgroup$ – mikuszefski May 10 '17 at 15:49
  • $\begingroup$ Yeah why? I am trying to say I didn't make a mistake. But if you use amperes law the result would be different. $\endgroup$ – most venerable sir May 10 '17 at 15:51
  • $\begingroup$ ...just wrote it because you say *Edit: Here is my professor's answer key to problem C. * $\endgroup$ – mikuszefski May 10 '17 at 15:54
  • $\begingroup$ Can you give a comparison of Biot-Savart vs Ampère, where do you get different results? They should be the same. They usually differ if you made a mistake and/or forgot about some additional effects. $\endgroup$ – mikuszefski May 10 '17 at 16:00
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Your basic problem here is thinking that these are different rules, when they are all aspects of a single rule. You just need to know the relationship between them: which are general and which apply to specific cases.

The Biot-Savart Law is the most general means of computing the magnetic field due to a sustained current distribution: it works for all problems. (Of course, the integral can be a bear in some cases.)

Ampere's law is also a general statement, but using it to find the magnetic field due to a current distribution is difficult, because what it tells use is the integral of field around a loop $\oint \vec{B} \cdot \mathrm{d}\vec{l}$, and it is only possible to extract a value of $B$ at a particular place from that in a few cases (either highly symmetric or with the field either perpendicular to the path or zero over much of the integration path).

The third formula you exhibit $$\mathbf{B = \frac{\mu_0}{2 \pi} \frac{I}{r} \tag{infinite line current}}$$ is a special case pertaining to the field surrounding a infinitely long uniform linear current (or a descent experimental approximation thereunto), and can be found from the Biot-Savart law by 'simply' integrating over the current distribution to which the special case applies (taking the current in the $\hat{z}$ direction through the origin, and the point of interest in the $z=0$ plane, and no return) \begin{align*} \vec{B}_\text{line current} &= \frac{\mu_0}{4\pi} \int_{-\infty}^\infty \frac{I \,\mathrm{d}\vec{z} \times \vec{r}}{(z^2 + r^2)^{3/2}} \\ \\ &= \frac{\mu_0}{4\pi} Ir \hat{\theta} \int_{-\infty}^\infty \frac{ \,\mathrm{d}z}{(z^2 + r^2)^{3/2}} \\ \\ &= \frac{\mu_0}{4\pi} Ir \frac{1}{r^2} \hat{\theta} \\ \\ &= \frac{\mu_0}{2\pi} \frac{I}{r} \hat{\theta} \\ \end{align*} or from Ampere's law with a rather easier computation (because this system has perfect rotational symmetry making it easy to extract meaning from Ampere's law.

You can't use the special case for problem (B) because those wires are neither very long, nor (for two for the segments) straight so the preconditions are not met. Further the system isn't neat enough to make Ampere's law attractive, so going with the Biot-Savart relationship is the only real choice.

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  • $\begingroup$ Ok when wire is infinitely long, you use ampere law. $\endgroup$ – most venerable sir May 10 '17 at 16:07
  • $\begingroup$ @user11355 ...you can use Ampère, but you don't have to. Biot-Savart just works as fine; this is the important message in the first two sentences of this answer. $\endgroup$ – mikuszefski May 10 '17 at 16:11
  • $\begingroup$ My takeaway from your answer is that the ampere shortcut is derived from BIOT SAVART by integrating over the infinite length of the wire. $\endgroup$ – most venerable sir May 10 '17 at 16:16
  • $\begingroup$ The part of Ampère's law I'd rephrase. It will always "work". If by "work" you mean determining $B$, however, low symmetry cases might be troublesome. The other way around, i.e. getting $I$ from the path integral surely "works". $\endgroup$ – mikuszefski May 10 '17 at 16:16
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    $\begingroup$ @user11355 Ampere's Law and the Biot-Suvart law carry roughly the same information as both are very general statement, but the Biot-Suvart Law gives the magnetic field at a particle point explicitly, while Ampere's law fiver you information about the field along a closed path rather than in one place. The $mu_o I/(2 \pi r)$ form, is a special case of either of the others. $\endgroup$ – dmckee --- ex-moderator kitten May 10 '17 at 16:47
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Problem $(c)$ is just a different version except in this case the magnetic field will the perpendicular to the plane containing the wires and through the point $"A"$.

The magnetic field due to a current carrying wire is given by :

$\vec{dB} = \frac{\mu_\circ}{4\pi}\int \frac{I\vec{dl}\space\times\space\hat{{r}}}{r^2}$

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$\vec{r}$ is the point where we want to find magnetic field due to the small current carrying element $\vec{dl}$.

In order to get the magnitude of the magnetic field at a point P, $a$ units away from the wire, we will have to integrate the above equation and, thus adding the magnetic field by all the small $dl$ elements in the current carrying wire of length L.

In problem (b), the integration procedure is the same. You add up the magnetic field due to all the $dl$ elements of the wire in such a way that when you integrate you have one o variable in the integral viz, '$\theta$'. You can convert length of arc into $\theta$ using the relation, $ length\space of\space Arc=R.\theta$ (R is the radius of the arc).

I'll skip the integration part, and on integrating we get to following formula,

$|\vec{B}|=\frac{\mu_\circ I}{4\pi a} [\sin{\phi_1}+\sin{\phi_2}]$

For infinite wires, $\sin{\phi_1}=\sin{\phi_2}=90^{\circ}$.

Thus, $|\vec{B}|=\frac{\mu_\circ I}{2\pi a}$

The direction of the magnetic field you can use the right hand thumb rule:

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Imagine holding the current carrying wire in your hand with your thumb pointing in the direction of the current. Now curl you finger around the wire. The fingers will encircle the wire in direction of the lines of magnetic force.

Remember that the magnetic field's direction at a point ,say a distance away, from a infinite wire is perpendicular to the plane containing the wire and the line joining the wire to the point where we want to find the magnetic field. You can determine it's direction as stated above.

In the problem $(c)$, magnetic field due to the wire towards left at point "A" will be upwards (out of the computer screen) as the current is going downwards and similarly you can find the direction and magnitude of magnetic field for the other wire.

At the point "C", the magnetic field's direction due to the wires will be opposite to each other.

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