A rotating cylinder in a magnetic field

From experience, if we spin a non-rigid body then it will expand. But most materials we encounter are rigid, and spinning them does not result in any noticeable expansion. This is because the internal forces become greater and provide the necessary centripetal force. The internal forces are generally chemical in nature, and chemical forces are essentially always electrostatic (dipole-dipole attraction of atoms/molecules or Coulomb attraction of ions). Thus, spinning a conductor would result in an internal electrical force to provide the centripetal acceleration.

But we are given that there is no internal electric field! Thus, the centripetal force must be provided by something else, and the only other thing it can be is the given magnetic field. Magnetic force act as follows: ${\bf F}_B = q {\bf v} \times {\bf B}$

We only care about magnitudes here, and the simple rotation allows us to state that $v = \omega r$ . The rotation also means that all the velocity is in the plane of rotation, so ${\bf v}$ is orthogonal to ${\bf B}$ . This means that $|{\bf v} \times {\bf B}| = v B$ . Finally, to put the centripetal statement above into symbols, we have: $F_{cent} = F_B$ . Putting this all together, we have: $m \omega^2 r = F_{cent} = F_B = q v B = q \omega r B$

Simplifying: $B = \frac{m \omega}{q}$

We note that the point particles whose $m$ and $q$ we desire are electrons, as those are the free charge carriers abundant in conductors. Thus: $B = \frac{m \omega}{q} = \omega \left(\frac{e}{m_e}\right)^{-1}$ $B = (1600 \ \mathrm{s}^{-1}) (1.76 \times 10^{11} \ \mathrm{C} / \mathrm{k g})^{-1} = \boxed{9.09 \times 10^{-9} \ \mathrm{T}}$

This is the desired answer.

In this problem, we must recall two facts about electrical conductors:

1. There is no electric field inside it;
2. Its electrons are free to move across its surface .

Since the charges will be in movement (because the cylinder is rotating) and will be in the presense of a magnetic field, a magnetic force will appear. And, by the left-hand rule, we know that it will be towards the cylinder's axis:

$F_1 = -ev \times B$ ,

where $v$ is the velocity of the electron and $e$ its charge. The minus signal indicates the force is towards the axis.

Let our frame move together with the cylinder. Hence, a centrifugal force will appears:

$F_2 = m_e v² /R$ .

But keep in mind that the material is originally neutral; and that its charges are kept in its surface. If, for instance, one electron leave the surface, there would be a resulting charge on the surface, and, therefore, a resulting electrical field from the cylinder. As we don't want this happens, no electrons can leave the surface. Therefore, the resulting force on each electron must be $0$ .

Therefore,

$F_1 + F_2 = 0$ ,

$evB=mv²/R$ ,

$eB=mv/r$ .

But because $v = \omega R$ , we have $eB = m \omega$ . Hence,

$B = \frac{m_e}{e}\omega$ .

Using the data given in the problem, we get finally

$B = \frac{1600}{1.76*10^{11}}$ $\approx 9.09*10^{-9}$ .

The conductor is originally neutral, implying that the electrons in it are all evenly distributed amongst the positive, immobile nuclei. However, the rotation creates an electric field, implying that this distribution is changed, creating regions of unequal charge, and ergo an electric field between them.

Indeed, the equation for centripetal force is $F = mr\omega^{2}$ , where $\omega$ is the angular velocity, $r$ is the radius and $m$ is the mass. This implies that the force on the electrons increases the further they are from the axis of rotation, and therefore that the electrons in the conductor are all being drawn towards the axis of rotation, causing the centre of the conductor to be negatively charged and the parts of it further from the centre to be positively charged. This sets up an electric field.

For the magnetic field to prevent the creation of this electric field, the Lorentz force it causes has to counteract the centripetal force. Since the equation for the Lorentz force is $F = Bqv$ , where $q$ is the charge, this gives us the following equation to work with: $Bqv = mr\omega^{2}$ .

However, we have a rogue $v$ in this equation, so to remove it, we substitute in $v = r\omega$ , which gives $Bqr\omega = mr\omega^{2}$ .

Cancelling $r\omega$ , we get $Bq = m\omega$ , which, fortunately, gets rid of the $r$ . Divide on both sides by $q$ , and we get $B = \frac{m}{q}\omega = (1.76 \times 10^{11})^{-1}(1600) = \boxed{9.09 \times 10^{-9}}$ , quod erat demonstrandum .

The two forces acting on the electrons are the centripetal force due to rotation and the magnetic force due to the external magnetic field. If there is no electric field created inside the conductor, the two forces should cancel. The centripetal force is given by $F_c=m_e\omega^2R$ where $m_e$ is the mass of the electron, $\omega$ is the angular velocity, and $R$ is the radius of the solid conductor.The magnetic force is given by $F_B=e\omega RB$ where $e$ is the charge of one electron and $B$ is the magnitude of the magnetic field. Equating these two, $e\omega RB=m_e \omega^2 R$ $\frac{eB}{m_e}=\omega$ $B=\frac{\omega}{e/m_e}$ Plugging the values in, $B=\frac{1600}{1.76\times 10^{11}}=\boxed{9.09\times 10^{-9}}$

Let the radius of the cylinder be $r$ , and let its rotational velocity be $\omega$ . Consider an electron on the outer shell of the cylinder. The linear velocity of this electron will be $v= \omega r$ . The forces acting on it are:
i) Outward centrifugal force with magnitude $\frac{mv^2}{r} = m \omega \times v$
ii) Inward Lorentz force with magnitude $evB$
Since the net electric force is zero, the electron must be at equilibrium, hence the magnitudes of these forces must be equal. Equating, we obtain: $evB= m \omega \times v$ $\implies B= \frac{m \omega}{e}$ Plugging in the values, we obtain $B \approx \boxed{9.09 \times 10^{-9} \text{ Teslas}}$ .