A particle spiraling in a magnetic field

Let the initial velocity of the particle before entering the magnetic field be v and let its velocity just after the magnetic field is switched off be u.Basically, we need to find v. Initial radius = R = frac{m * v}{B * q} Final radius = 0.98 * R = frac{m * u}{B * q} Thus, dividing the two equations, we get u = 0.98 * v Given that drag force F = m a = - k v , where v is velocity of the particle at that instant. We know that a = v * frac{dv}{dx}
Integrating the above equation with v varying from v to u and x varying from 0 to 4 \pi R , we get m (u-v) = - k * 4 \pi R. Substituting u = 0.98 * v, we get frac{k}{m} = frac{0.02 * v}{4 \pi R}. Now, integrate the force equation with velocity varying from u to 0 and distance x varying from 0 to L, we get frac{k}{m} = frac{u}{L}. Substituting the values of u, L and frac{k}{m} in this equation, we can find R to be equal to 0.000487 metres. But, R = frac{m * v}{B * q} Thus, substituting the values R, B and frac{q}{m}, we get v = 487.2 m/s.

From the equality of the magnetic centripetal force against the centrifugal force we get: For initial entry velocity Vi and exit velocity Ve from the chamber: q/m=Vi/(RixB)=Ve/(RexB), where Ri-initial radius and Re exit radius and given Ri/Re=1/0.98=1.02. Now: Fdrag=-KxV and a(acceleration)=dv/dt=-KxV/m or dt=(dV/V(-K/m)) and set m/K=z integrating on both sides will give Ve=Vixexp(-t/z) The way the particle makes in the chamber before stopping the Mag. field is approx 4xPi(Ri+Re)/2 or two complete rotations with an average radius of AVG{Ri,Re}. On the other hand the distance can be computed from the integral of the momentary speeds around the circles: S=INT{VixExp(-t/z)} which equals to 4xPi(Ri+Re)/2 All I get Vixz(1-exp(-t/z))=4xPi(Ri+Re)/2 assuming the lower time limit=0 it is not hard to see that we get Vixz-Vexz=4Pix(m/q)x(1/B)(Vi+Ve)/2 when using the relations between the radii to the speeds. Transfer it to the form 1.02xVexz-Vexz where 1.02=1/0.98 Vexz=The stopping distance in our case 0.3meter which we get from the equation S=Vexzx(1-exp(-t/z)) by sustitution of t=infinite or S=Vexz=0.3meter in this case From now on just a hard long work to place the different items in place and get the final solution. finally for short Vexz(or 0.3)*(1.02-1)=4xPix(m/q)x(1/B)xVix(1+0.98)x(1/2) out of which we can compute Vi the speed of the particle when entering the chamber.

motion of the particle after switching off the magnetic field:

m \frac {dv}{dt} = - k v

by the solution of the differential equation we obtain

v {t} = v {0} e^(-kt/m) (1)

\frac {dx}{dt} = v_{0} e^(-kt/m)

by solving the differential equation again, we obtain

x = - v \frac {m}{k e^(kt/m)} + v \frac {m}{k}

when the particle comes to stop , time is infinity

substituting obtains v_{0} = 0.3 \frac {k}{m} (2)

motion while rotating:

qvB = m \frac {v^2}{r}

by substituting we obtain the following:

in the beginning of rotation : r = \frac {v_{0}}{10^6}

in the end of rotation : r = \frac {v {t}}{r(1-2%)} = \frac {v {t}}{r \times 0.96*10^6}

\frac {v_{0}}{v(t)} = \frac {1}{0.98)}

while \frac {qB}{m} = \frac {v}{r}

then, the angular velocity is always constant

t = \frac {theta}{omega}

t = \frac {4 * pi}{10^6} = 1.256637 \times 10^-5 second

using equation (1), but while rotating, we obtain

v {t} = v {0} e^(-kt/m) (3)

\frac {v_{0}}{v(t)} = e^(kt/m)

\frac {1}{0.98)} = e^(\frac {1}{0.98)} * 1.256637 \times 10^-5)

by solving the equation we obtain

\frac {k}{m} = 1607.6804

we know that v {0} in equation (2) equals v {t} in equation (3)

so we obtain

0.3 \frac {k}{m} = v_{0} e^(-kt/m)

after solving the equation we can get the value of v_{0}

v_{0} = 492.147 m/s

Since the radius of curvature changes gradually we can write $m \frac{v^{2}}{R}= q v B \rightarrow R= \frac{v}{\alpha B}.$ Therefore, a small change of the particle's velocity ( $\Delta v$ ) implies a small change of the radius R ( $\Delta R$ ). From the last equation we have that $\frac{\Delta R}{R}=\frac{\Delta v}{v}\approx \frac{\Delta{v}}{v_{0}} \rightarrow \frac{\Delta{v}}{v_{0}}=-\epsilon$ where $\epsilon= 2/100$ . Thus the velocity of the particle after two rotations is $v_{1}= v_{0} (1-\epsilon).$ When the magnetic field is switched off, Newton's second Law assumes the form $m \frac{dv}{dt}= -k v \rightarrow m \frac{dv}{dt}=- k \frac{ds}{dt} \rightarrow m\int dv =- k \int ds$ which implies $v_{1}=\frac{k}{m} L \rightarrow v_{0} (1-\epsilon) =\frac{k}{m} L.$ In order to solve the problem we need an additional equation. Note that $m \frac{dv}{dt}= -k v \rightarrow m\int dv =- k \int ds$ can also be applied when the particle is spiraling inward. Then it yields $m \Delta v = - k s = -k 4\pi R$ and since $R= \frac{v}{\alpha B}$ we arrive at the relation $\epsilon =\frac{ k}{m} \frac{4\pi}{\alpha B}$ Now we can get rid of the unknown ratio $\frac{k}{m}$ to obtain $v_{0}=\frac{\alpha B\epsilon L}{4\pi (1-\epsilon)}=487.2 \textrm{m/s}.$