# Force on non-moving charge in solenoid with increasing magnetic field

The problem: "Through a solenoid there is an increasing magnetic field $B$. For a positively charged particle not in motion, what is the direction of force on the particle?"

Normally a non-moving particle in a magnetic field would equal no charge. Here, however, there is an increasing magnetic field into the page. This would induce an emf providing a current moving against the clock. By Lenz's law there would be a new increasing magnetic field opposing the original one. This is where I'm at a loss - how would this affect the charge? The $z$-direction is parallel to the magnetic fields, but both $y$ and $x$ is normal.

$\newcommand{\curl}[1]{\nabla \times #1}\newcommand{\pdv}[2]{\frac{\partial #1}{\partial #2}}$There is an emf created because of the Lenz's Law, which means an electric field is being created. The electric field is circular (thus the force, which is proportional to the electric field, is also circular) and the charged particle would spiral out due to added velocity by the circular force. Depending on the charge of the particle and the direction of the increasing magnetic field, it would go either clockwise or anti-clockwise.

$$\curl {\vec E} = - \pdv {\vec B} t$$

and assuming that $\vec B(t)= \beta t \hat z$ for some constant $\beta$, where $\hat z$ is the unit vector along the $z$ direction, we have (I'll drop the vector arrows from now on)

$$(\curl E)_z = - \beta \quad, \tag 1$$

where $(\curl E)_z$ denotes the $z$ component of $\curl E$.

Changing to the cylindrical coordinates $(r, \theta, z)$ we have,

$$(\curl E )_z = \frac 1 r \left( \pdv {(r E_\theta)} r - \pdv {E_r} \theta \right) = \frac {E_\theta} r + \pdv {E_\theta} r - \frac 1 r \pdv {E_r} \theta \quad. \tag 2$$

From the symmetry of the problem it is clear that $\partial E_r/ \partial \theta$ is zero. Therefore substituting equation $(2)$ in the Faraday's law $(1)$ we are left with the differential equation

$$\pdv {E_\theta} r + \frac {E_\theta} r + \beta = 0$$

One can take the ansatz$^*$ $\tilde E_\theta = \frac {C_1} r - C_2 \beta x$ for some constants $C_1$ and $C_2$. We have

$$\pdv {\tilde E_\theta} r + \frac {\tilde E_\theta} r + \beta = \beta - 2\beta C_2 = 0 \implies C_2= \frac 1 2 \quad .$$

Hence we can write the real solution as

$$E_\theta = \frac C r - \beta \frac r 2$$

for some constant $C$ that should be determined by the initial conditions. Note that when $C \neq 0$ funny things happen near the origin, in particular the electric field diverges. Now we can avoid this problem by two ways:

1. We'll look at the solution, which is far a way from the origin, i.e. as $r \to \infty$ the term $C/r \to 0$. Therefore we will proceed by ignoring that term altogether.
2. We'll say that the solution for $C\neq 0$ is not physical at the origin, therefore we'll set $C=0$.

You can choose whichever reason you want but the upshot is that we'll ignore the $C/r$ term.

Then we have for the electric field

$$\fbox{ E_\theta = - \beta \frac r 2 }\quad. \tag 3$$

Easily enough the force on the particle is given by

$$\vec F = q \vec E = - \beta \frac {q \cdot r} 2 \, \hat \theta \quad ,$$

where $\hat \theta$ denotes the unit vector in the $\theta$ direction, which you can convert into Cartesian coordinates by substituting

$$\hat \theta = \begin{pmatrix} -\sin \theta \\ \cos \theta \\ 0 \end{pmatrix} \quad \text {with} \quad \theta = \arctan \frac y x$$

and remembering that $r= \sqrt{x^2+y^2}$.

In any event you know that $\theta = \rho$ for some constant $\rho$ give a cylinder in cylindrical coordinates. Thus it is trivial to see that the solution $(3)$ describes a circular electric field, which in return corresponds to an outgoing spiral motion of the particle as we predicted using only our intuition.

$^*$As to why we have chosen that ansatz i.e. an educated guess as to why the solution should look like that, we first start by noting that

$$\pdv {} r \frac 1 r = - \frac 1 {r^2} = \frac 1 r \cdot \frac 1 r$$

Hence a term of the form $C_1/r$ takes care of the term $E_\theta/r$ up to a constant term, so we must include $C_1/r$ in our ansatz $\tilde E_\theta$.

Having taken care of the term $E_\theta/r$ we are left with a pesky constant $\beta$. My intuition says that adding a term of the form $C_2 \beta r$ should clean up the $\beta$ when differentiated with respect to $r$. However adding this term may also screw up the $E_\theta / r$ term. Luckily however what we want to add depends only on $r$ in the first order, hence differentiating this term with respect to $r$ and dividing by $r$ gives the same result! For safety one always adds a dummy constant $C_2$ in front of this term, which you should adjust later on, which is exactly what I did above.

• Unless the particle is constrained to remain at a radius r from the centre, I don't think the particle would actually move in a circle if it's in the circular electric field - to do that it would need to accelerate towards the centre of the solenoid, not tangentially. – Chris Cundy Jul 31 '15 at 13:42
• @ChrisCundy: At the first glance I thought it should stay on a circle but I think about it more, it think it should spiral out due to the added velocity by the circular electric field. – Gonenc Mogol Jul 31 '15 at 13:45
• Easier solution is to use Stokes's theorem and equate the line integral of the magnetic vector potential to the flux through the loop and then take the time derivative to get the electric field. Just assume cylindrical symmetry. No Ansatz required. You haven't explained why $E_r=0$. – Rob Jeffries Oct 26 '18 at 10:03

In a time-independent field you get a circular motion with constant speed. This is due to the Lorentz force perpendicular to the velocity vector of the particle. If the magnetic field strength is increasing however, an electrical field is induced due to the increasing flux that the loop created by the moving particle catches. This electrical field will increase the speed of the particle and, through the Lorentz force, diminish the radius of circular orbit of the particle. This mechanism is used to heat plasmas (adiabatic compression of the plasma).

Excellent explanation on page 35-36 of this document

• Whilst this may theoretically answer the question, it would be preferable to include the essential parts of the answer here, and provide the link for reference. – Gonenc Mogol Jul 31 '15 at 12:59