Gauss Law for Magnetic fields?

When facing a physics problem, the first rule is do no harm.

Therefore, since we're told this field has a radial component, let us not pick a tetrahedron or a cube as a test surface to evaluate the flux integral. Instead, let us go with the cylindrically symmetric cylinder.

On the first accusation, we can easily see that a $z$ -component only field would violate the no-flux condition.

Say for sake of argument that the test cylinder is 2 units high (from $z=-1$ to $z=1$ ), and let's denote the normal vectors to the surfaces $\hat{A}$ . Then the flux at the top is equal to

$\hat{A}\cdot\vec{B} = \pi R^2\left(B_0 + b\right)$

and at the bottom is given by

$\hat{A}\cdot\vec{B} = \pi R^2\left(-B_0 + b\right)$

for an overall flux of $2\pi R b$ piercing the wall, which is not zero.

The only surface on the cylinder that can save us with a negative compensating flux is the wall. Hence, the radial field is required.

Let us say there is a radial field and moreover that it points inwards (an outwards facing field will make more positive flux which is just further from our goal of zero flux).

Moreover let us assume the cylinder to have arbitrary height, $z$ . Our results cannot depend on the geometry of our arbitrarily chosen test surface and therefore had best not depend on the arbitrary height of the arbitrary shape of our arbitrary test surface.

The flux through the top and bottom is given by $2\pi R b$ as we showed above. Through the wall we have a field $B_r$ penetrating the wall which has a total surface area $2\pi R\cdot 2z$ making a flux contribution of $-4\pi R z B_r$

We want these two fluxes to sum to zero, therefore:

$\displaystyle 4\pi R z B_r = 2\pi R b$

and we have

$\boxed{\displaystyle B_r = \frac{Rb}{2} = 2.5\mbox{E}-7\mbox{ T}}$

Consider a cylindrical surface parallel to the z-axis, with radius 50cm. The two bases of the cylinder are at z=0 and z=t, respectively.

Total magnetic flux through the bases = Area x B = (pi r^2)(B0 + bt - B0) = (pi r^2)bt

Thus there must be an inward magnetic flux through the side of the cylinder, so that the total magnetic flux becomes 0

Let Br be the magnitude of the radial component. Then (Br)(2pi r) = (pi r^2)bt Br = rb/2 = 2.5E-7 T

For magnetostatic systems, the divergence $\nabla . \mathbf{B}$ of $\mathbf{B}$ must be $0$ . In cylindrical coordinates, so that $\mathbf{B} \,=\, (B_r,B_\theta,B_z)$ , this means that $0 \; = \; \frac{1}{r}\frac{\partial}{\partial r}(rB_r) + \frac{1}{r}\frac{\partial B_\theta}{\partial \theta} + \frac{\partial B_z}{\partial z}$ In our case, this means that $\frac{1}{r}\frac{\partial}{\partial r}(rB_r) + b \; = \; 0$ and hence $rB_r \,=\, \alpha - \tfrac12br^2$ , and hence $B_r \,=\, \alpha r^{-1} - \tfrac12br$ . Since we want $B_r$ to be continuous, we must have $\alpha=0$ , and hence $B_r = -\tfrac12br$ .

When $r=0.5$ , we obtain $B_r = -2.5 \times 10^{-7}$ T.

Imagine a cylinder with radius $R$ and length $2z$ (along the $z$ axis), let the centre of mass of the cylinder be the origin, therefore

The magnetic flux through the top of the cylinder is $\pi R^2 (B_0+bz)$

The magnetic flux through the bottom of the cylinder is $\pi R^2 (-B_0+bz)$ , because the direction of the field $B_0$ is towards the opposite direction at the bottom

Adding them up, magnetic flux along $z$ axis is $\pi R^2 *2bz$

Since $\Phi _ M = 0$ ,

The radial component must be able to compensate for the flux along $z$ axis

The flux in radial direction (i.e through the walls of the cylinder) is the area of the rectangle times the strength of the magnetic field, $2\pi R * 2z * B_r$ where $B_r$ denotes the magnetic field along the radial direction

Equating the two fluxes,

$\pi R^2 *2bz = 2\pi R * 2z * B_r$

$B_r = \frac{Rb}{2} = \frac{0.5m \times 10^{-6}T/m}{2} = \boxed{2.5*10^{-7} T}$

Consider a cylinder with radius $r$ , height $dz$ and centered at the z-axis. Let the base of the cylinder be at height $z$ From Gauss' Law, total magnetic flux entering and exiting this cylinder is 0. Hence, $\Phi_{side} + \Phi_{top} + \Phi_{bottom}= 0$

$\Phi_{top} = -\pi r^2 \cdot (B_0 + b(z+dz))$ and $\Phi_{bottom} = \pi r^2 \cdot (B_0 + bz)$

By symmetry, $\Phi_{side} = B_r \cdot 2 \pi r$

Therefore, $-\pi r^2 \cdot (B_0 + b(z+dz)) + \pi r^2 \cdot (B_0 + bz) + B_r \cdot 2 \pi r = 0$

$B_r \cdot 2 \pi r = dz \cdot \pi r^2 \implies B_r = \frac{b \cdot r}{2} = \frac{1 \times 10^{-6} \cdot 0.5}{2} = 2.5 \times 10^{-7}$