91影视

Finding the expressions for three magnetic fields.

To get the magnetic field of Ampere's law need to integrate over the circumference of the enclosed area.

$$\begin{gathered} {鈭�}_0^l\stackrel{鈫�}{B}路\stackrel{鈫�}{dI}={渭}_o{I}_{encl} \\ B{\left[l\right]}_0^{2蟺r}={渭}_o{I}_{encl} \\ B\left(2蟺r\right)={渭}_o{I}_{encl} \\ B=\frac{{渭}_o{I}_{encl}}{2蟺r} \end{gathered}$$

The current density inside the enclosed area (r) equals the current density in the whole wire of radius $\left(R_1\right)$

$J_1=\frac{1}{蟺R_1^2}$

$I$is the current of wire.

$J_r=\frac{I_{encl}}{蟺r^2}$

The current$I_{encl}$

$$\begin{gathered} J_1=J_r \\ \frac{I}{蟺R_1^2}=\frac{I_{encl}}{蟺r^2} \\ {I}_{encl}=\frac{r^2}{R_1^2}I \end{gathered}$$

Need to get B with the expression of I into equation (1).

$B=\frac{{渭}_o{I}_{encl}}{2蟺r}=\frac{{渭}_o\left({\displaystyle \frac{r^2}{R_1^2}}\right)}{2蟺r}=\boxed{\frac{{渭}_{o}Ir}{2蟺R_1^2}}$

The magnetic field is inside the wire when $r鈮�R_{1}whereB鈭�r$

The distance $R_2鈮�r鈮�R_1$and the current is the same for the wire with the radius

The distance$R_2鈮�r鈮�R_1$and the current is the same for the wire with the radius $R_1$

$$\begin{gathered} {鈭�}_0^l\stackrel{鈫�}{B}路\stackrel{鈫�}{dI}={渭}_{o}I \\ B{\left[l\right]}_0^{2蟺r}={渭}_{o}I \\ B\left(2蟺r\right)={渭}_{o}I \\ \boxed{B=\frac{{渭}_I}{2蟺r}} \end{gathered}$$

$r鈮�R_2$

The current flows at Distance $r鈮�R_2$the net current in the path is zero. From Ampere's law the magnetic field will be zero

$$\begin{gathered} 鈭�\stackrel{鈫�}{B}路\stackrel{鈫�}{dI}={渭}_o\left(0\right) \\ \boxed{B=0} \end{gathered}$$

In the magnetic field the surface of the inner wire decreases as r increases till reach$R_2$, after this point it becomes zero.