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If the magnetic field of an infinite wire at a distance $R$ is is $0.4 \;\text{T}$, find the field at a distance $2R$.

The approach suggested by internet is as follows:

Magnetic field due to an infinite wire is: $$B = \frac{\mu_0I}{2\pi R}$$ or as given: $$0.4 = \frac{\mu_oI}{2\pi R}$$

When R is doubled, $$B = \frac{\mu_0I}{2\pi (2R)}$$

On simplification, the answer comes out to be $0.2 \;\text{T}$. It relies on the fact that the angle is not much changed since the wire is relatively infinite, which seems fine to me. But is the following method correct to do the same question?

$$0.4 = \frac{\mu_0I}{4\pi} \int \frac{\text{d}l \sin\theta}{R^2} \tag{1}$$

When R is doubled, $$\begin{align}B = \frac{\mu_0I}{4\pi} \int \frac{\text{d}l \sin\theta}{(2R)^2} \\ = \frac{\mu_0I}{16\pi} \int \frac{\text{d}l \sin\theta}{R^2} \tag{2}\end{align}$$

Dividing (1) by (2),

$$\frac {0.4}{B} = \left(\frac{\mu_0I}{4\pi} \int \frac {\text{d}l \sin\theta}{R^2}\right) / \left(\frac{\mu_0I}{16\pi} \int \frac {\text{d}l \sin\theta}{R^2}\right) \\ \frac {0.4}{B} = 4 \\ B = 0.1 \;\text{T}$$

If the angle is not affected due to the fact that the wire is infinite, then the integration part should get cancelled. Please help me with this.

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    $\begingroup$ No, the second method is incorrect because the $\text{d}l$'s are not the same. When you double the distance, $\text{d}l$ also doubles for the same $\text{d}\theta$. $\endgroup$ Sep 23 at 13:22
  • $\begingroup$ @VincentThacker the dl is the snall length of wire which remains same in both the cases $\endgroup$ Sep 23 at 13:52
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    $\begingroup$ No, you are integrating theta, not $l$, from -90 degrees to +90 degrees. Try drawing a picture and you will see what I mean $\endgroup$ Sep 23 at 13:57
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First case:

enter image description here

Second case:

enter image description here

You seem to be confused to between $r$ and $R$. $r$, used in the Biot-savart law, is the distance of all the elements along the wire from point, whereas $R$ is the distance of the midpoint of the wire from the point.

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The fact is that the 'R' you are using in the first simplified equation is the perpendicular distance of the point from the wire. Whereas the R you are using in the integration is not the same , the R used here is the distance of the point from the end point of the infinite wire , which is infinite.Here, see the image below for a small wire... enter image description here The thing you are confusing is that you are thinking R and r represented in the image as same.

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In the integration, r is the distance of the elementary part from the concerned dl part. So you cammot directly put that 2R, as this r is something that depends on theta, and is 2R only for theta= 90 degress. So it is recommended to use the first one, as in that r is the perpendicular distance from the line to the concerned point.

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  • $\begingroup$ But in both cases angle is considered 90°. Whats the problem. $\endgroup$ Sep 23 at 13:53
  • $\begingroup$ I don't see how that is relevant? $\endgroup$
    – satan 29
    Sep 23 at 14:28
  • $\begingroup$ In the integration, the r is varying as the angle varies, but in the final formula R, the one mentioned at the beginning, is basically the perpendicular distance of the point from the wire. So you can directly change that. But if you want that to be made directly in the integration, then you have to adjust the integration accordingly. $\endgroup$
    – Rhitam
    Sep 24 at 6:13

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