Magnetic field from a wire at a point

[Decimal]

Homework Statement

Picture of the problem:

I first need to calculate the magnetic field in point P caused by the piece of wire at z = R (so the top wire parallel to the y-axis).

Homework Equations

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Biot-Savart law for magnetic fields:

$\vec {dB} = \frac {μ0*I} {4*π} * \frac {\vec {dl} ×\vec {r}} {r^3}$

The Attempt at a Solution

[/B]
Here is what I did:

$\vec {dl} = dy * \vec {ey}$ (where ey is the unit vector in the y direction)
$\vec r = (P - y)* \vec {ey} - R * \vec {ez}$

then:

$\vec {dl} ×\vec r = -R*dy*\vec {ex}$

so now I can write:

$\vec {dB} = - \frac {μ0*I*R} {4*π} *\frac {1} {(P - y)^2 + R^2)^{3/2}} * dy * \vec {ex}$

Now I can integrate this using the substitution $u = P-y$ and $du = -dy$, and then integrate from $-L$ to $0$, which gives me an expression for the magnetic field:

$\vec B =\frac {μ0*I} {4*π*R} *(\frac {P} {\sqrt {P^2+R^2}} -\frac {P+L} {\sqrt {(P+L)^2+R^2}}) * \vec {ex}$

Now this answer is almost correct except there should be a minus in front. However my minus disappeared when I used the substitution $du = -dy$. Does anyone know where I went wrong?

 

[kuruman]

Check the details of the integration. You probably used a tangent substitution. How is the positive angle defined?

 

[Decimal]

Integration:

$\vec B =- \frac {μ0*I*R} {4*π} * \int \frac {1} {(P - y)^2 + R^2)^{3/2}} * \vec {ex} \, dy$

$\vec B = - \frac {μ0*I*R} {4*π} * \int \frac {1} {(u)^2 + R^2)^{3/2}} * \vec {ex} \,* -du$

I am allowed to use the solution to some standard integrals, like this one, so I can just write out the solution directly:

$\vec B = - \frac {μ0*I*R} {4*π} * [\frac {-1} {R^2} * \frac {u} {\sqrt {u^2+R^2}}] \vec {ex}$

Then I substitute $u = P-y$ back in and evaluate the indefinite integral from $-L$ to $0$ which gets me to my final answer mentioned above:

$\vec B =\frac {μ0*I} {4*π*R} *(\frac {P} {\sqrt {P^2+R^2}} -\frac {P+L} {\sqrt {(P+L)^2+R^2}}) * \vec {ex}$

This is what I did, as you can see my minus sign canceled out. I am not sure what you mean by the positive angle.

 

[kuruman]

Why is the z-component of $\vec{r}$ negative?

 

[Decimal]

$\vec r$ should point from the line segment $\vec {dl}$ to point $P$ right? So then it would have a negative z-component since it should point from $z=R$ to $z=0$

 

[kuruman]

Yes, you're right. I think you put one minus sign too many with dy and the limits of integration. If you use the the right hand rule and the diagram below, you get
$\vec {dB} = - \frac {μ_0*I*R} {4*π} *\frac {1} {[(P + y)^2 + R^2]^{3/2}} * dy * \hat {x}$
Note the positive sign in the denominator. Also note that the limits of integration should be from zero to L, not negative L because the smallest value for the y-component of $\vec{r}$ is P and the largest value is P + L.

 

[Decimal]

But shouldn't $r = \sqrt {(P-y)^2+P}$? Since y is always negative in this case so the y component of $\vec r = P-y$

 

[kuruman]

I think your problem is with the back substitution from $u$ to $y$.
You have the indefinite integral (after cancellation of the two negative signs)
$\vec B = \frac {μ_0I} {4πR} \int \frac {du} {(u^2+R^2)^{3/2}} \hat {x} = \frac{u}{R^2\sqrt{R^2+u^2}}$
Since $u=P-y$, when $y=0$, $u = P$ and when $y=-L$, $u=P+L$. These are the lower and upper limits that you should put back in the indefinite integral for $u$.

Edited to clarify points and fix mistakes.

 

[Decimal]

Yeah I think I have figured where I am going wrong, though I don't understand why yet. I did perform the back substitution of the limits as you mention in your post, however I am integrating from -L to 0, whilst you are integrating from 0 to -L. I thought you always have to integrate in the direction of the unit vector? Or am I misunderstanding this?

 

[kuruman]

I thought you always have to integrate in the direction of the unit vector?

Not necessarily. It is element dy that is always assumed to be positive. The limits of integration add a negative sign or not.

 

[Decimal]

mmm okay, so how do I know what to use as integration limits then? In this case dy points in the same direction as the current, so I figured I should just integrate in that direction too, since its also the direction of the unit vector. How would I know to integrate in the other direction in this case?

Thank you for your help by the way

 

[kuruman]

Ignore the unit vector and its direction. The key things are the integrand, f(q) dq and the limits of integration. Note that I used q, a dummy variable of integration that eventually disappears when you evaluate the integral. You should always bear in mind that when you integrate, the "something" in d(something} can be any symbol. I think your confusion arises because you are treating y as both a dummy variable of integration and a Cartesian coordinate. In post #6 I have y increasing to the left. Strictly speaking, I should have called it something else, say q, because y should be reserved for the Cartesian coordinate that increases to the right.

Look at the drawing you made when you worked on this problem. Call the distance along the y-axis of element $d \vec{l}$ q, set up the definite integral and see where it takes you.

 

[TSny]

$\vec B =\frac {μ0*I} {4*π*R} *(\frac {P} {\sqrt {P^2+R^2}} -\frac {P+L} {\sqrt {(P+L)^2+R^2}}) * \vec {ex}$

Now this answer is almost correct except there should be a minus in front.

Your answer looks correct to me. Note that your answer gives a negative number times the unit vector e_x. This is the correct direction for B. I don't think there should be a minus sign in front.

 

[Decimal]

Alright I found a similar exercise in with more elaborate explanations and they also don't arrive at a minus sign, so I think the answer is just wrong, like TSny said. Thanks a ton for your help though Kuruman since I have a much better understanding of how to perform the integration now, though even with your tips I would still arrive at my original answer.

Thanks!

 

[kuruman]

I don't think there should be a minus sign in front.

Right you are. When I first looked at this, I concentrated on the denominator and ignored the $P+L$ in the numerator.