91Ó°ÊÓ

Two wires, parallel to Z the axis and distance 4r apart carry equal current i in opposite directions.

Hint: The flux through the portion of the xz plane that lies within the cylinder.

Applying the Gauss law for magnetism, the flux through the S₁portion, i.e., the upside portion, is equal to the flux through the S₂, i.e., downside portion, of the XY plane that lies within the cylinder in terms of magnetic field and length of a circular cylinder. Then using the formula for the magnetic field due tothecurrent passing through an infinite straight wire, the expression can be foundfor the net outward magnetic flux through half of the cylindrical surface above the x-axis.

The formula is as follows:

${\mathit{Φ}}_{\mathbf{B}}\mathbf{=}\mathbf{âˆ\circledR }\stackrel{\mathbf{â‡\P Ä}}{\mathbf{B}}\mathbf{.}\mathit{d}\stackrel{\mathbf{â‡\P Ä}}{\mathbf{A}}\mathbf{.}$

Where,

${\mathrm{Φ}}_B$is the Magnetic flux

B is the magnetic induction

A is the area

Applying the Gauss law for magnetism, the flux through the S₁ portion, i.e., the upside portion, is equal to the flux through the S₂, i.e., downside portion, of the xz plane.

According to Gauss law,

${\mathrm{Φ}}_{\mathrm{B}}=âˆ\circledR \stackrel{â‡\P Ä}{\mathrm{B}}.d\stackrel{â‡\P Ä}{\mathrm{A}}.$

In this case, the magnetic flux through S₁ portion is localid="1663069281307" ${\mathrm{Φ}}_B\left(S_1\right)=\underset{-r}{\overset{r}{∫}}B\left(x\right)Ldx$.

Similarly, the magnetic flux through portion S₁ is localid="1663069373927" ${\mathrm{Φ}}_B\left(S_2\right)=\underset{-r}{\overset{r}{∫}}B\left(x\right)Ldx$.

But the magnitude of the magnetic field isthesame in both portions. So,

${\mathrm{Φ}}_B\left(S_1\right)={\mathrm{Φ}}_B\left(S_2\right)$

The net magnetic flux is,

${\mathrm{Φ}}_B\left(S\right)=\underset{-r}{\overset{r}{∫}}B\left(x\right)Ldx$

where the magnetic field due to the current passing through an infinite straight wire is,

$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}r}$

Here, the r is 2r - x

So thatthemagnetic field is,$B=\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}\left(2r-x\right)}$.

Hence,

$\begin{array}{rcl}{\mathrm{Φ}}_B\left(s\right)& =& 2\underset{-r}{\overset{r}{∫}}\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}\left(2r-x\right)}Ldx\\ {\mathrm{Φ}}_B\left(s\right)& =& \frac{{\mathrm{μ}}_{0}iL}{\mathrm{π}}\underset{-r}{\overset{r}{∫}}\frac{dx}{\left(2r-x\right)}\\ {\mathrm{Φ}}_B\left(s\right)& =& \frac{{\mathrm{μ}}_{0}iL}{\mathrm{π}}{\left[-In\left(2r-x\right)\right]}_{-r}^r\\ {\mathrm{Φ}}_B\left(s\right)& =& \frac{{\mathrm{μ}}_{0}iL}{\mathrm{π}}\left[In\left(3r\right)-In\left(r\right)\right]\\ {\mathrm{Φ}}_B\left(s\right)& =& \frac{{\mathrm{μ}}_{0}iL}{\mathrm{π}}\left[In\left(\frac{3r}{r}\right)\right]\\ {\mathrm{Φ}}_B\left(s\right)& =& \frac{{\mathrm{μ}}_{0}iL}{\mathrm{π}}In3\end{array}$

Hence the expression for the net outward magnetic flux through half of the cylindrical surface above the x-axis is,${\mathrm{Φ}}_B\left(s\right)=\frac{{\mathrm{μ}}_{0}iL}{\mathrm{π}}In3$.

Using the concept of Gauss law for magnetism, the expression for the net outward magnetic flux through half of the cylindrical surface above the x-axis can be found.