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Question: Write an expression for the magnetic field not in the center of the loop and not even at a point lik...

Write an expression for the magnetic field not in the center of the loop and not even at a point like this on a convenient axis but at an arbitrary point in space. Do deepthink and write it in details step by step, do deep research for it. It is an upper class level advanced problem

Answer

The expression for the magnetic field B⃗(x,y,z)\vec{B}(x,y,z) at an arbitrary point (x,y,z)(x,y,z) due to a circular loop of radius RR lying in the xyxy-plane and carrying a current II is: B⃗(x,y,z)=μ0IR4π∫02πzcos⁡ϕ′i^+zsin⁡ϕ′j^+(R−xcos⁡ϕ′−ysin⁡ϕ′)k^(x2+y2+z2+R2−2R(xcos⁡ϕ′+ysin⁡ϕ′))3/2dϕ′\vec{B}(x,y,z) = \frac{\mu_0 I R}{4\pi} \int_0^{2\pi} \frac{z\cos\phi' \hat{i} + z\sin\phi' \hat{j} + (R - x\cos\phi' - y\sin\phi') \hat{k}}{(x^2+y^2+z^2+R^2 - 2R(x\cos\phi' + y\sin\phi'))^{3/2}} d\phi'

Explanation

Solution

The fundamental principle used to calculate the magnetic field produced by a steady current is the Biot-Savart Law. For a small segment of wire carrying current II, dl⃗d\vec{l}, the magnetic field dB⃗d\vec{B} at an observation point with position vector r⃗\vec{r} due to this segment located at r⃗′\vec{r}' is given by: dB⃗(r⃗)=μ04πI dl⃗′×(r⃗−r⃗′)∣r⃗−r⃗′∣3d\vec{B}(\vec{r}) = \frac{\mu_0}{4\pi} \frac{I \, d\vec{l}' \times (\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3} To find the total magnetic field B⃗(r⃗)\vec{B}(\vec{r}) at the observation point, we integrate this expression over the entire current loop CC: B⃗(r⃗)=∮Cμ04πI dl⃗′×(r⃗−r⃗′)∣r⃗−r⃗′∣3\vec{B}(\vec{r}) = \oint_C \frac{\mu_0}{4\pi} \frac{I \, d\vec{l}' \times (\vec{r} - \vec{r}')}{|\vec{r} - \vec{r}'|^3} For a circular loop of radius RR in the xyxy-plane centered at the origin, with current II, we parameterize the loop using the angle ϕ′\phi'. A point on the loop is r⃗′(ϕ′)=Rcos⁡ϕ′i^+Rsin⁡ϕ′j^\vec{r}'(\phi') = R \cos\phi' \hat{i} + R \sin\phi' \hat{j}, and the differential length vector is dl⃗′=Rdϕ′(−sin⁡ϕ′i^+cos⁡ϕ′j^)d\vec{l}' = R d\phi' (-\sin\phi' \hat{i} + \cos\phi' \hat{j}). The observation point is r⃗=xi^+yj^+zk^\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}. The vector difference is r⃗−r⃗′=(x−Rcos⁡ϕ′)i^+(y−Rsin⁡ϕ′)j^+zk^\vec{r} - \vec{r}' = (x - R\cos\phi')\hat{i} + (y - R\sin\phi')\hat{j} + z\hat{k}. The magnitude squared is ∣r⃗−r⃗′∣2=(x−Rcos⁡ϕ′)2+(y−Rsin⁡ϕ′)2+z2=x2+y2+z2+R2−2R(xcos⁡ϕ′+ysin⁡ϕ′)|\vec{r} - \vec{r}'|^2 = (x - R\cos\phi')^2 + (y - R\sin\phi')^2 + z^2 = x^2+y^2+z^2+R^2 - 2R(x\cos\phi' + y\sin\phi'). The cross product dl⃗′×(r⃗−r⃗′)d\vec{l}' \times (\vec{r} - \vec{r}') is calculated to be: dl⃗′×(r⃗−r⃗′)=Rdϕ′[zcos⁡ϕ′i^+zsin⁡ϕ′j^+(R−xcos⁡ϕ′−ysin⁡ϕ′)k^]d\vec{l}' \times (\vec{r} - \vec{r}') = R d\phi' [ z\cos\phi' \hat{i} + z\sin\phi' \hat{j} + (R - x\cos\phi' - y\sin\phi')\hat{k} ] Substituting these into the Biot-Savart integral and integrating from ϕ′=0\phi'=0 to 2π2\pi yields the expression for the magnetic field at an arbitrary point (x,y,z)(x,y,z). This integral is complex and typically evaluated using special functions like complete elliptic integrals.