Question

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.

Short answer

Answer

The expression for the electric and magnetic field are

$$\begin{gathered} \frac{q}{4\pi {\epsilon }_0}\frac{Rc}{{\left(Rc\right)}^2}\left\{\left[\left({\omega }^2{R}^2-c^2\right)\cos \left(\omega t_r\right)+\omega Rc\sin \left(\omega t_r\right)\right]\hat{x} \\ +\left[\left({\omega }^2{R}^2-c^2\right)\sin \left(\omega t_r\right)+\omega Rc\cos \left(\omega t_r\right)\right]\hat{y} \\ \right\} \end{gathered}$$ and $\frac{q}{4\pi {\epsilon }_0}\frac{\omega }{R{c}^2}\hat{z}$respectively. The expression for magnetic field is at the centre of circular loop is as same as the Ex 5.6 with in equation $B\left(z\right)=\frac{I{\mu }_0}{2}\frac{R^2}{{\left(R^2+z^2\right)}^{\frac{3}{2}}}$.

Step-by-step solution

Write the given data from the question.

The steady state current in the circular loop is I.

The charge moves in circular loop of radius r .

Determine the formula to calculate the electric and magnetic field at the centre.

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.

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

Calculate the electric and magnetic field at the centre.

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

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