Question

Suppose you wanted to find the field of a circular loop (Ex. 5.6) at a point $r$that is not directly above the center (Fig. 5.60). You might as well choose your axes so that $r$lies in the $yz$plane at $(0,y,z)$. The source point is ( $R$cos ${\phi }^{\text{'}},R$sin $\left.{\varphi }^{\text{'}},0\right)$, and ${\varphi }^{\text{'}}$runs from 0 to $2J$J. Set up the integrals25 from which you could calculate $B_x,B_y$and $B_z$and evaluate $B_x$explicitly.

Short answer

The $x,y$ and $z$ component of the magnetic field of the circular loop are 0

Step-by-step solution

Determine the position vector and its magnitude for magnetic field.

Consider the figure for the given condition:

The position vector for the magnetic field is given as:

$\hat{p}=-R\cos \varphi +(y-R\sin \varphi )\hat{y}+z\hat{z}$

Here, $\varphi$is the angle for the circular loop and $R$is the radius of the circular loop.

The magnitude of the position vector is given as:

$\backslash begin\left\{aligned\right\}\&p=\backslash sqrt\left\{\right(R\backslash \cos \backslash phi)^\{2\}+(y-R\backslash \sin \backslash phi)^\{2\}+z^\{2\left\}\right\}\backslash \backslash \&p=\backslash left\left[\right(R\backslash \cos \backslash phi)^\{2\}+(y-R\backslash \sin \backslash phi)^\{2\}+z^\{2\}\backslash right]^\{1/2\}\backslash end\left\{aligned\right\}$

The length of the small element of the circular loop is calculated as:

$\backslash begin\left\{aligned\right\}\&l\_\{\backslash bar\{y\left\}\right\}^\left\{z\right\}=(R\backslash \cos \backslash phi)\backslash hat\left\{x\right\}+(R\backslash \sin \backslash phi)\backslash hat\left\{y\right\}+z\backslash hat\left\{z\right\}\backslash \backslash \&dl\_\left\{y\right\}=-(R\backslash \sin \backslash phi)d\backslash phi\backslash hat\left\{x\right\}+(R\backslash \cos \backslash phi)d\backslash phi\backslash hat\left\{y\right\}+0\backslash \backslash \&dl=-(R\backslash \sin \backslash phi)d\backslash phi\backslash hat\left\{x\right\}+(R\backslash \cos \backslash phi)d\backslash phi\backslash hat\left\{y\right\}\backslash end\left\{aligned\right\}$

The vector product of position vector and length vector of circular loop is given as:

$\backslash begin\left\{aligned\right\}\&d\backslash vec\left\{d\right\}\backslash times\backslash hat\left\{p\right\}=[-(R\backslash \sin \backslash phi)d\backslash phi\backslash hat\{x\}+(R\backslash \cos \backslash phi)d\backslash phi\backslash hat\{y\left\}\right]\backslash times[-R\backslash \cos \backslash phi\backslash hat\{x\}+(y-R\backslash \sin \backslash phi)\backslash hat\{y\}+z\backslash hat\{z\left\}\right]\backslash \backslash \&dl\backslash times\backslash hat\left\{p\right\}=(Rz\backslash \cos \backslash phid\backslash phi)\backslash hat\left\{x\right\}+(Rz\backslash \sin \backslash phid\backslash phi)\backslash hat\left\{y\right\}+\backslash left(-Ry\backslash \sin \backslash phid\backslash phi+R^\{2\}+d\backslash phi\backslash right)\backslash hat\left\{z\right\}\backslash end\left\{aligned\right\}$

Determine the components of magnetic field of circular loop

The x component of the magnetic field of circular loop is given as:

$B_x=\frac{{\mu }_0{l}^{2\pi }}{4\pi }{\int }_0^{2\pi }\left(\frac{dl\times \stackrel{\Delta }{p}}{p^3}\right)$

Substitute all the values in the above equation.

$\backslash begin\left\{aligned\right\}\&B\_\left\{x\right\}=\backslash frac\{\backslash mu\_\{0\}I^\{2\backslash pi\left\}\right\}\{4\backslash pi\}\backslash int\_\left\{0\right\}^\left\{2\right\}\backslash frac\{\backslash left[(Rz\backslash \cos \backslash phid\backslash phi)\backslash hat\left\{x\right\}+(Rz\backslash \sin \backslash phid\backslash phi)\backslash hat\left\{y\right\}+\backslash left(-Ry\backslash \sin \backslash phid\backslash phi+R^\{2\}+d\backslash phi\backslash right)\backslash hat\left\{z\right\}\backslash right\left]\right\}\{\backslash left[\backslash left\left[\right(R\backslash \cos \backslash phi)^\{2\}+(y-R\backslash \sin \backslash phi)^\{2\}+z^\{2\}\backslash right]^\{1/2\}\backslash right]^\{3\left\}\right\}\backslash \backslash \&B\_\left\{x\right\}=0\backslash end\left\{aligned\right\}$

The y component of the magnetic field of circular loop is given as:

$B_y=\frac{{\mu }_0}{4\pi }{\int }_0^{2\pi }\left(\frac{d{l}^{\frac{L}{R}\times p}}{p^3}\right)$

Substitute all the values in the above equation.

$B_y=\frac{{\mu }_0}{4\pi }{\int }_0^{2\pi }\frac{\left[\left(Rz\cos \varphi d\varphi \right)\hat{x}+\left(Rz\right)\hat{y}+\left(-Ry\sin \varphi d\varphi +R^2+d\varphi \right)\hat{z}\right]}{{\left[{\left[(R\cos \varphi {)}^2+(y-R\sin \varphi {)}^2+z^2\right]}^{1/2}\right]}^3}$

$B_y=\frac{{\mu }_{0}IR{z}^{2\pi }}{4\pi }{\int }_0^2\frac{\sin \varphi d\varphi }{{\left[(R\cos \varphi {)}^2+(y-R\sin \varphi {)}^2+z^2\right]}^{3/2}}$

The$z$ component of the magnetic field of circular loop is given as:

$B_z=\frac{{\mu }_0{J}^{2\pi }}{4\pi }{\int }_0^N\left(\frac{dl\times p^N}{p^3}\right)$

Substitute all the values in the above equation.

$\backslash begin\left\{aligned\right\}\&B\_\left\{z\right\}=\backslash frac\{\backslash mu\_\{0\}I^\{2\backslash pi\left\}\right\}\{4\backslash pi\}\backslash int\_\left\{0\right\}^\left\{\right[\}\backslash frac\{[Rz\backslash \cos \backslash phid\backslash phi)\backslash hat\left\{x\right\}+(Rz)\backslash hat\left\{y\right\}+\backslash left(-Ry+R^\{2\}+d\backslash phi\backslash right)\left\}\right\{\backslash left[\backslash left[(R\backslash \cos \backslash phi)^\left\{2\right\}+(y-R\backslash \sin \backslash phi)^\left\{2\right\}+z^\left\{2\right\}\backslash right]^\{1/2\}\backslash right]^\left\{3\right\}\}\backslash \backslash \&B\_\{z\}=\backslash frac\{\backslash mu\_\left\{0\right\}IR^\{2\backslash pi\}\left\}\right\{4\backslash pi\}\backslash int\_\{0\}^\{2\backslash pi\}\backslash frac\{(R-y\backslash \sin \backslash phi)d\backslash phi\left\}\right\{\backslash left\left[\right(R\backslash \cos \backslash phi)^\{2\}+(y-R\backslash \sin \backslash phi)^\{2\}+z^\{2\}\backslash right]^\{3/2\}\}\backslash end\{aligned\}$

Therefore, the $\mathrm{x},\mathrm{y}$and $\mathrm{z}$component of the magnetic field of the circular loop are 0 ,

$\backslash begin\left\{aligned\right\}\&\backslash frac\{\backslash mu\_\{0\}IRz\}\{4\backslash pi\}\backslash int\_\left\{0\right\}^\{2\backslash pi\}\backslash frac\{\backslash \sin \backslash phid\backslash phi\}\{\backslash left[(R\backslash \cos \backslash phi)^\left\{2\right\}+(y-R\backslash \sin \backslash phi)^\left\{2\right\}+z^\left\{2\right\}\backslash right]^\{3/2\left\}\right\}\backslash text\{and\}\backslash \backslash \&\backslash frac\{\backslash mu\_\{0\}IR\}\{4\backslash pi\}\backslash int\_\left\{0\right\}^\{2\backslash pi\}\backslash frac\left\{\right(R-y\backslash \sin \backslash phi)d\backslash phi\}\{\backslash left[(R\backslash \cos \backslash phi)^\left\{2\right\}+(y-R\backslash \sin \backslash phi)^\left\{2\right\}+z^\left\{2\right\}\backslash right]^\{3/2\left\}\right\}\backslash end\left\{aligned\right\}$