For the configuration in Prob. 10.15, find the electric and magnetic fields at the center. From your formula for B, determine the magnetic field at the center of a circular loop carrying a steady current I, and compare your answer with the result of Ex. 5.6.

The steady state current in the circular loop is I.

The charge moves in circular loop of radius r .

The expression for the position of charge at any point is given as follows.

$W\left(t\right)=R\left[cos\left(\omega t\right)\hat{x}+sin\left(\omega t\right)\hat{y}\right]$

Here,W (t) is the position of charge, is the time and is the angular velocity.

The expression to calculate the velocity is given as follows.

$v\left(t\right)=\omega R\left[-sin\left(\omega t\right)\hat{x}+cos\left(\omega t\right)\hat{y}\right]$

The expression to calculate the velocity is given as follows.

$a\left(t\right)=-{\omega }^{2}W\left(t\right)$

The expression for electric field due to moving charge is given as follows.

$E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r\times u\right)}^3}\left[\left(c^2-v^2\right)u+r\times \left(u\times a\right)\right]$

Here, q is the charge, v is the speed, c and is the velocity of light.

The expression for the magnetic field is given as follows.

$B=\frac{1}{c}\left(\hat{r}\times E\right)$

Consider the figure shown below,

It is known that:

r = R

The points can be calculated as,

$$\begin{gathered} t_r=t-\frac{R}{c} \\ \hat{r}=-\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{x}\right] \end{gathered}$$

Calculate the expression for as,

$u=\hat{r}-v\left(t_r\right)$

Substitute$-C\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]$for $\hat{r}$and $\omega R-C\left[-\sin \left(\omega t_r\right)\hat{x}+\cos \left(\omega t_r\right)\hat{y}\right]$for into above equation.

Calculate the value of as,

$$\begin{gathered} r·a=-W\left(-{\omega }^{2}W\right) \\ r·a={\omega }^2{R}^2\left({\cos }^2\omega t+{\sin }^2\omega t\right)\backslash \\ r·a={\omega }^2{R}^2\backslash \end{gathered}$$

Calculate the value of as,

$$\begin{gathered} r·u=R\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]\left\{\left[c.\cos \left(\omega t_r\right)-\omega R\sin \left(\omega t_r\right)\hat{x}\right]+\left[c.\cos \left(\omega t_r\right)+\omega R\sin \left(\omega t_r\right)\hat{x}\right]\right\} \\ r·u=R\left\{\left[\left[{\cos }^2\left(\omega t_r\right)\hat{x}+-\omega R\cos \left(\omega t_r\right)\sin \left(\omega t_r\right)+\right]\right]\left[{\sin }^2\left(\omega t_r\right)\hat{x}+-\omega R\sin \left(\omega t_r\right)\cos \left(\omega t_r\right)\right]\right\} \\ r·u=R\left[{\cos }^2\left(\omega t_r\right)+{\sin }^2\left(\omega t_r\right)\hat{x}\right] \\ r·u=R \end{gathered}$$Calculate the expression for the electric filed.

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+r\times \left(u\times a\right)\right] \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{r}{{\left(r·u\right)}^3}\left[\left(c^2-v^2\right)u+\left(r·a\right)u-\left(r·u\right)a\right] \end{gathered}$$

Substitute Rc for r.u and wr for r.a into equation (4).

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{R}{{\left(Rc\right)}^3}\left[\left(c^2-{\omega }^2{R}^2\right)u+\left({\omega }^2{R}^2\right)u-\left(Rc\right)a\right] \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{R}{{\left(Rc\right)}^3}\left(c^{2}u-\left(Rc\right)a\right) \end{gathered}$$

Substitute $\left[c.\cos \left(\omega t_r\right)-\omega R\sin \left(\omega t_r\right)\right]\hat{x}+\left[c\sin \left(\omega t_r\right)-\omega R\cos \left(\omega t_r\right)\right]$for u , and ${\left(-R\omega \right)}^2\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]$for a into above equation,

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{R}{{\left(Rc\right)}^3}\left\{c^2\left[\cos \left(\omega t_r\right)+-\omega R\sin \left(\omega t_r\right)\right]\hat{x}+\left[c.\sin \left(\omega t_r\right)+-\omega R\cos \left(\omega t_r\right)\right]-Rc-R{\omega }^2\left[\cos \left(\omega t_r\right)\hat{x}+\sin \left(\omega t_r\right)\hat{y}\right]\right\} \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{Rc}{{\left(Rc\right)}^2}\left\{\left[\left({\omega }^2{R}^2-c^2\right)\cos \left(\omega t_r\right)++\omega Rc\sin \left(\omega t_r\right)^\right]\hat{x}\left[\left[\left({\omega }^2{R}^2-c^2\right)Sin\left(\omega t_r\right)++\omega Rc\cos \left(\omega t_r\right)^\right]\hat{y}\right]\right\} \\ E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{1}{{\left(Rc\right)}^2}\left\{\left[\left({\omega }^2{R}^2-c^2\right)\cos \left(\omega t_r\right)++\omega Rc\sin \left(\omega t_r\right)^\right]\hat{x}\left[\left[\left({\omega }^2{R}^2-c^2\right)Sin\left(\omega t_r\right)++\omega Rc\cos \left(\omega t_r\right)^\right]\hat{y}\right]\right\} \end{gathered}$$

Calculate the expression for the magnetic field,

Substitute $E\left(r,t\right)=\frac{q}{4\pi {\epsilon }_0}\frac{1}{{\left(Rc\right)}^2}\left\{\left[\left({\omega }^2{R}^2-c^2\right)\cos \left(\omega t_r\right)++\omega Rc\sin \left(\omega t_r\right)\right]\left[\left[\left({\omega }^2{R}^2-c^2\right)Sin\left(\omega t_r\right)++\omega Rc\cos \left(\omega t_r\right)^\right]\hat{z}\right]\right\}$for into equation (5).

$$\begin{gathered} B=\frac{1}{C}\left(\begin{array}{c}\frac{q}{4\pi {\epsilon }_0}\frac{1}{{\left(Rc\right)}^2}\left[\left({\omega }^2{R}^2-c^2\right)\cos \left(\omega t_r\right)++\omega Rc\sin \left(\omega t_r\right)\right]\hat{x}\\ +\left[\left({\omega }^2{R}^2-c^2\right)Sin\left(\omega t_r\right)++\omega Rc\cos \left(\omega t_r\right)\right]\hat{y}\end{array}\right) \\ B=-\frac{q}{4\pi {\epsilon }_0}\frac{1}{R^2{c}^3}\left(-\omega Rc{\sin }^2\left(\omega t_r\right)-\omega Rc{\cos }^2\left(\omega t_r\right)\hat{z}\right) \\ B=\frac{q}{4\pi {\epsilon }_0}\frac{1}{R^2{c}^3}\omega Rc\hat{z} \end{gathered}$$ …….. (1)

Hence the expression for the electric and magnetic field are

$B=\frac{q}{4\pi {\epsilon }_0}\frac{1}{R^2{c}^3}\left(\begin{array}{c}\left({\omega }^2{R}^2-c^2\right)\cos \left(\omega t_r\right)++\omega Rc\sin \left(\omega t_r\right)\\ \left({\omega }^2{R}^2-c^2\right)\sin \left(\omega t_r\right)++\omega Rc\cos \left(\omega t_r\right)\end{array}\right)$

Determine the magnetic field at the centre of the ring charge.

The ring carries the I current , therefore,

$$\begin{gathered} I=\lambda v \\ I=\lambda \omega R \\ \lambda =\frac{I}{\omega R} \end{gathered}$$

The expression for the charge is given by,

$q=\lambda \left(2\pi R\right)$

Substitute $\frac{I}{\omega R}$for into above equation.

$$\begin{gathered} q=\frac{I}{\omega R}\left(2\pi R\right) \\ q=2\pi \frac{I}{\omega } \end{gathered}$$

Substitute$2\pi \frac{I}{\omega }$ for q into equation (1).

$$\begin{gathered} B=\frac{2\pi I}{4\pi \omega {\epsilon }_0}\frac{\omega }{R{c}^2}\hat{z} \\ B=\frac{I}{2{\epsilon }_{0}R{c}^2}\hat{z} \end{gathered}$$

Substitute${\mu }_0{\epsilon }_0$ for$\frac{1}{c^2}$ into above equation.

$$\begin{gathered} B=\frac{I{\mu }_0{\epsilon }_0}{2{\epsilon }_{0}R}\hat{z} \\ B=\frac{I{\mu }_0}{2R}\hat{z} \end{gathered}$$

The expression for magnetic field is same as the Ex 5.6 with in equation

$B\left(z\right)=\frac{I{\mu }_0}{2}\frac{R^2}{{\left(R^2+z^2\right)}^{\frac{3}{2}}}$