Figure 32-27 shows a closed surface. Along the flat top face, which has a radius of 2.0 cm, a perpendicular magnetic field $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{B}}$of magnitude 0.30 T is directed outward. Along the flat bottom face, a magnetic flux 0.70 mWb is directed outward. What are the (a) magnitude and (b) direction (inward or outward) of the magnetic flux through the curved part of the surface?

• The radius of the top face is R = 2.0 cm = 0.02 m.
• The magnetic field out of the top face is B_Top = 0.30 T.
• Magnetic flux directed outward through the bottom face${\phi }_{bottom}=0.70mWb$

Write an equation for the net magnetic flux through the given closed surface using Gauss law for magnetism. Then inserting the given values, find the magnitude and direction of the magnetic flux through the curved part.

The formula is as follows:

${\phi }_b=\oint B.dA$

In the given closed surface, apply Gauss law, i.e.,

Net magnetic flux through any closed surface is zero.

$$\begin{gathered} {\phi }_b=\oint B.dA=0 \\ {\phi }_b={\phi }_{top}+{\phi }_{bottom}+{\phi }_{curvedsurface}=0...............\left(1\right) \end{gathered}$$

It is known,${\phi }_{bottom}=0.70mWb=0.70\times {10}^{-3}Wb$

$$\begin{gathered} {\phi }_{top}=\oint B.dA=B_{top}{\mathrm{\pi r}}^2 \\ {\mathrm{\phi }}_{\mathrm{top}}=0.30\times 3.14\times 0.{02}^2 \\ {\mathrm{\phi }}_{\mathrm{top}}=0.0003768\mathrm{Wb} \\ {\mathrm{\phi }}_{\mathrm{top}}=0.37\times {10}^{-3}\mathrm{Wb} \end{gathered}$$

Using equation (1),

$\begin{array}{rcl}0.37\times {10}^{-3}+0.70\times {10}^{-3}+{\phi }_{\phi curvedsurface}& =& 0\\ 1.07\times {10}^{-3}+{\phi }_{\phi curvedsurface}& =& 0\\ {\phi }_{\phi curvedsurface}& =& -1.07\times {10}^{-3}\\ {\phi }_{\phi curvedsurface}& =& 1.1mWb\\ \left|{\phi }_{\phi curvedsurface}\right|& =& 1.1mWb\end{array}$

Hence, the magnitude of the magnetic flux through the curved surface is 1.1 mWb .

The direction of magnetic flux through the curved surface is inwards because its sign is negative.