Electrons held by magnetic field

[tan90]

Homework Statement

Electrons are held by a magnetic field B in circular orbit in a vacuum chamber. The electrons are acclerated by increasing the magnetic flux linking the orbit. Prove that the average magnetic field over the plane of the orbit must be twice the magnetic field at the orbit if the orbit is to remain fixed as the electron's energy is increased.

Homework Equations

F_centripetal = mv²/r
F_magnetic = qvBe

The Attempt at a Solution

F_centripetal = F_magnetic
mv²/r = qvB
1/2 (mv²) = qvBr
K.E. = qvBr

I don't know if the approach is right, I am doing self study for this chapter. I would appreciate some help.

 

[tan90]

oh i messed up a little bit above,
1/2 mv^2 = 1/2 qvBr
2K.E. = qvBr

 

[jdstokes]

I haven't been able to figure this out but I think I have the right idea.

As the magnetic field increases, it induces an electric field which does work on the electron, increasing its velocity. Since the magnetic field at the orbit is increasing, the radius stays constant since

$r = mv/eB$.

Appying Faraday's law, I find that after an infinitesimal period of time, the velocity increases by dv = (e/m)(r/2) dB.

Hmm.

 

[jdstokes]

I'm actually getting that B ~ r^2, which implies that the average is half that at the orbit. Are you sure there isn't a mistake in the question?

 

[tan90]

that looks like a right idea to me...I checked the question, it should be twice not half at the orbit and even if you got B~ r^2, how is it going to be half ;$, i am confused.

 

[jdstokes]

Well, if the the magnetic field inside the orbit is less than or equal to the value at the orbit, this implies that the average value is less than or equal to the value at the orbit.

If the average is greater than the value at the orbit, the magnetic field must be decreasing as you move away from the center of the loop.

 

[jdstokes]

I'll show you what I've done. According to Faraday's law

$\oint \vec{E}\cdot d\vec{\ell} = - \frac{d}{dt} \int \vec{B} \cdot d\vec{a}.$

Integrating around the orbit gives

$2\pi r E = - \frac{d}{dt} 2\pi\int_{0}^r B(r') r' dr' \implies$
$rEdt = - \int_{0}^r \partial B(r') r' dr'.$

Assuming that $dB(r) = dB$ then

$rEdt = - dB \frac{r^2}{2} \implies dv = \frac{e}{m}\frac{r}{2}dB$

but $v = \frac{eBr}{m} \implies dv = \frac{eB}{m}dr$

so

$\frac{eB}{m}dr = \frac{e}{m}\frac{r}{2} dr \implies 2 \frac{dr}{r} = \frac{dB}{B} \implies$

$B\propto r^2$.

 

[tan90]

my question seems to be wrong plus it is logical only if B over the orbit is half the B at the orbit. thanks a bunch.

 

[jdstokes]

The question is correct. It's an adaptation of a question out of Griffiths.

First differentiate the expression qvB = mv^2/r to obtain

E = r dB/dt.

The rest you can show using the integral form of Faraday's law around the electron orbit.