91Ó°ÊÓ

• The radius of the caps of the cylinder r = 12.0 cm = 0.12 m
• Length of the cylinder L = 80 cm = 0.8 m
• At the first cap 1, inward magnetic flux, ${\mathrm{φ}}_1=-24\mathrm{μ}Wb=-25×{10}^{-6}Wb$
• At the other cap 2, the magnetic field$B_2=1.60mT=1.60×{10}^{-3}T$

Applying Gauss law for magnetism, write for magnetic flux through thecircular cylinder. Inserting given values in it, find the magnitude and direction of the magnetic flux through the curved part.

The formula is as follows:

Gauss law for any closed Gaussian surface,${\mathrm{φ}}_b=âˆ\circledR B.dA=0$

In the given closed surface, that is, the circular cylinder, G ${\mathrm{φ}}_{cap2}=0.0723456×{10}^{-3}Wb$. Gauss law gives that net magnetic flux through any closed surface is zero.

$$\begin{gathered} {\mathrm{φ}}_b=âˆ\circledR B.dA=0 \\ {\mathrm{φ}}_b={\mathrm{φ}}_{cap1}+{\mathrm{φ}}_{cap2}+{\mathrm{φ}}_{curvedsurface}=0..........\left(1\right) \end{gathered}$$

It is known,${\mathrm{φ}}_{cap1}=-24×{10}^{-6}Wb$

$$\begin{gathered} {\mathrm{φ}}_{cap2}=âˆ\circledR B.dA=B_{cap2}{\mathrm{Ï€°ù}}^2 \\ {\mathrm{φ}}_{\mathrm{cap}2}=1.60×{10}^{-3}×3.14×{\left(0.12\right)}^2 \\ {\mathrm{φ}}_{\mathrm{cap}2}=72.4×{10}^{-6}\mathrm{Wb} \end{gathered}$$

Using this in equation 1,

role="math" localid="1663055685557" $\begin{array}{rcl}-25×{10}^{-6}+72.4×{10}^{-6}+{\mathrm{φ}}_{curvedsurface}& =& 0\\ 47.4×{10}^{-6}+{\mathrm{φ}}_{curvedsurface}& =& 0\\ {\mathrm{φ}}_{curvedsurface}& =& -47.4×{10}^{-6}Wb\\ {\mathrm{φ}}_{curvedsurface}& =& -47.4\mathrm{μ}Wb\\ \left|{\mathrm{φ}}_{curvedsurface}\right|& =& 47.4\mathrm{μ}Wb\end{array}$

Therefore, the magnitude of magnetic flux through the curved surface is $\begin{array}{rcl}& & 47.4\mathrm{μ}Wb\end{array}$.

The direction of magnetic flux through a curved surface is inwards since its negative.

Applying Gauss law for magnetism, find the magnetic flux through one part using the magnetic flux through the other parts of the closed surface.