In Fig. 32-43, a bar magnet lies near a paper cylinder. (a) Sketch the magnetic field lines that pass through the surface of the cylinder. (b) What is the sign of $\overrightarrow{\mathbf{B}}\mathbf{\cdot }\mathit{d}\overrightarrow{\mathbf{A}}$for every $\mathit{d}\overrightarrow{\mathbf{A}}$area on the surface? (c) Does this contradict Gauss’ law for magnetism? Explain.

The magnetic field is $\overrightarrow{\mathrm{B}}$.

You can find the angle between the cross-sectional area and magnetic field. Using the dot product of vectors, you determine the sign of the
$\overrightarrow{\mathbf{B}}\mathbf{\cdot }\mathit{d}\overrightarrow{\mathbf{A}}$.

Formula:

Use the vector formula as below.

$\overrightarrow{A}.\overrightarrow{B}=\left|A\right|\left|B\right|cos\theta$

Sketch the magnetic field lines which pass through the surface of the cylinder.

From the above diagram, we can say that at a point on the surface, normal to $d\overrightarrow{A}$is in the opposite direction to the magnetic field $\overrightarrow{B}$.

From this, we can say that the angle between area and the field will be $180°$. Therefore you have,

$\begin{array}{rcl}\overrightarrow{B}.d\overrightarrow{A}& =& \left|B\right|\left|dA\right|\cos \theta \\ & =& \left|B\right|\left|dA\right|\cos 180°\\ & =& -\left|B\right|\left|dA\right|\end{array}$

So, from the above equation, we can say that sign of $\overrightarrow{B}\cdot d\overrightarrow{A}$for every area$d\overrightarrow{A}$on the surface will be negative.

Gauss law for magnetism is applied to an enclosed surface. If we consider top and bottom portion of the cylinder to create an enclosed surface, we will get $\underset{}{\overset{}{}}\overrightarrow{B}·d\overrightarrow{A}=0$.

This is because the flux entering the surface would be equal to the flux leaving the surface.

This concludes that the result in part b) does not contradict Gauss law for magnetism