91影视

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.

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

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.

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

The expression to calculate the velocity is given as follows.

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

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

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

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.

$B=\frac{1}{c}\left(\hat{r}脳E\right)$

Consider the figure shown below,

It is known that:

r = R

The points can be calculated as,

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

Calculate the expression for as,

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

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

Calculate the value of as,

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

Calculate the value of as,

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

$$\begin{gathered} E\left(r,t\right)=\frac{q}{4蟺{蔚}_0}\frac{r}{{\left(r路u\right)}^3}\left[\left(c^2-v^2\right)u+r脳\left(u脳a\right)\right] \\ E\left(r,t\right)=\frac{q}{4蟺{蔚}_0}\frac{r}{{\left(r路u\right)}^3}\left[\left(c^2-v^2\right)u+\left(r路a\right)u-\left(r路u\right)a\right] \end{gathered}$$

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

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

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

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

Calculate the expression for the magnetic field,

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

$$\begin{gathered} B=\frac{1}{C}\left(\begin{array}{c}\frac{q}{4蟺{蔚}_0}\frac{1}{{\left(Rc\right)}^2}\left[\left({蝇}^2{R}^2-c^2\right)\cos \left(蝇t_r\right)++蝇Rc\sin \left(蝇t_r\right)\right]\hat{x}\\ +\left[\left({蝇}^2{R}^2-c^2\right)Sin\left(蝇t_r\right)++蝇Rc\cos \left(蝇t_r\right)\right]\hat{y}\end{array}\right) \\ B=-\frac{q}{4蟺{蔚}_0}\frac{1}{R^2{c}^3}\left(-蝇Rc{\sin }^2\left(蝇t_r\right)-蝇Rc{\cos }^2\left(蝇t_r\right)\hat{z}\right) \\ B=\frac{q}{4蟺{蔚}_0}\frac{1}{R^2{c}^3}蝇Rc\hat{z} \end{gathered}$$ 鈥︹��.. (1)

Hence the expression for the electric and magnetic field are

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

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

The ring carries the I current , therefore,

$$\begin{gathered} I=位v \\ I=位蝇R \\ 位=\frac{I}{蝇R} \end{gathered}$$

The expression for the charge is given by,

$q=位\left(2蟺R\right)$

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

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

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

$$\begin{gathered} B=\frac{2蟺I}{4蟺蝇{蔚}_0}\frac{蝇}{R{c}^2}\hat{z} \\ B=\frac{I}{2{蔚}_{0}R{c}^2}\hat{z} \end{gathered}$$

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

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

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

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