Electron's U-Turn

@Neeraj Anand Badgujar @Steven Chase

Okay, after a lot of struggle, here it is:

Consider a long wire of circular cross-section area of radius 2 placed along the Y-axis. The electron is projected from the point $(2,0,0$ with an initial velocity of $\vec{v}_o=v_o \hat{i}=1000 \hat{i}$ . The current flowing through the wire is along the positive Y direction.

The magnetic field at any general point in space is:

$\vec{B} = -\frac{\mu_o I}{2 \pi x} \hat{k}$

At any general instant of time, let the velocity of the particle be:

$\vec{V} = v_x \hat{i} + v_y \hat{j}$

Applying Newton's second law:

$m\vec{a} = -e\left(\vec{V} \times \vec{B}\right)$

Simplifying gives:

$\ddot{x} = \left(\frac{\mu_o Ie}{2 \pi m}\right) \frac{\dot{y}}{x}$ $\ddot{y} = -\left(\frac{\mu_o Ie}{2 \pi m}\right) \frac{\dot{x}}{x}$

Let:

$A = \frac{\mu_o Ie}{2 \pi m}$

This gives:

$\frac{dv_x}{dt} = \frac{Av_y}{x}$ $\frac{dv_y}{dt} = -\frac{Av_x}{x}$

Diving both equations above gives:

$\frac{dv_x}{dv_y} = -\frac{v_y}{v_x}$ $\implies v_x \ dv_x = -v_y \ dv_y$

Integrating and applying initial conditions gives:

$v_x^2 + v_y^2 = v_o^2$

This means that the speed of the particle is constant. Initially $v_y=0$ and we need to find the x-coordinate of the particle when $v_x$ becomes zero, as that would indicate a 'turning point'.

Using this intuition in the following equation:

$\frac{dv_y}{dt} = -\frac{Av_x}{x} = -\frac{A}{x}\frac{dx}{dt}$ $\implies dv_y = -\frac{A \ dx}{x}$

Recognising that as x increases, the y-coordinate decreases. To ensure this, an additional negative sign is introduced above which transforms the ODE to:

$dv_y = -\frac{A \ dx}{x}$

Integrating:

$\int_{0}^{-v_o}dv_y = A \int_{2}^{x_o} \frac{dx}{x}$

$\implies \frac{v_o}{A} = \ln\left(\frac{x_o}{2}\right)$

$\implies \boxed{x_o = 2 \mathrm{e}^{v_o/A}}$

I have reservations about the problem statement which I considered to be very misleading. This took me down a path where I unnecessarily over-complicated the problem.

The statement: 'From the surface of a round wire of radius 2m...' made me imagine that the wire is actually bent into the shape of a circular ring and that the wire has a negligible cross-section area. This confused me. I do not like poorly worded problems. Now I realise that others may find it easy to understand, but I am not one of those others, unfortunately. I require a great deal of clarity in statements.

I propose the following problem statement.

'Consider an infinitely long wire having a circular cross-section of radius 2. The wire is placed along the Y-axis and a current of $0.12 \ A$ flows uniformly throughout its cross-section along the positive Y direction (The distribution is actually irrelevant). An electron is projected from the surface of this long wire with a speed 0f 1000 m/s and its direction of projection is radially outward. Compute the maximum x-coordinate of the particle while it is in motion. Assume that the motion of the electron is affected only by the magnetic field generated by the wire.'

It would also help if you supplemented your problem statements with diagrams. I humbly re-iterate that you need to work on making your problem statements clearer. It may be clear to others but not for me. Thank you for understanding. I say this because I enjoy solving your problems and I wish to continue doing so.

Nice problem. I solved by running a numerical simulation through time. Code is attached, with comments:

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import math

dt = 10.0**(-9.0)  # time step

R = 2.0        # wire radius
I = 0.12       # wire current
v0 = 1000.0    # initial speed

q = 1.602*(10.0**-19.0)   # electron charge
m = 9.11*(10.0**-31.0)    # electron mass
u0 = 4.0*math.pi * (10.0**-7.0)  # magnetic permeability

#print (q/m)

#######################################

# Initialize simulation
# Assume wire is perpendicular to xy plane (along z axis)
# Initial electron velocity is along x axis

t = 0.0
count = 0

x = 2.0    # position
y = 0.0
z = 0.0

xd = v0    # velocity
yd = 0.0
zd = 0.0

xdd = 0.0  # acceleration
ydd = 0.0
zdd = 0.0

rsqd = 2.0*x*xd + 2.0*y*yd  # rate of change of squared distance from wire axis

##############################################

while rsqd > 0.0:   # run simulation while distance from wire axis increasing

    x = x + xd*dt    # numerical integration
    y = y + yd*dt
    z = z + zd*dt

    xd = xd + xdd*dt
    yd = yd + ydd*dt
    zd = zd + zdd*dt

    r = math.hypot(x,y)         # distance from wire axis
    rsqd = 2.0*x*xd + 2.0*y*yd  # rate of change of squared distance from wire axis

    B = u0*I/(2.0*math.pi*r)   # B field magnitude at charge location

    ux = x/r    # unit vector from wire axis to charge
    uy = y/r

    u2x = -uy   # unit vector in direction of B field at charge location
    u2y = ux    

    Bx = B*u2x  # vector magnetic flux density
    By = B*u2y
    Bz = 0.0

    crossx = yd*Bz - zd*By     # cross product for magnetic force  
    crossy = -(xd*Bz - zd*Bx)
    crossz = xd*By - yd*Bx

    Fx = q*crossx   # magnetic force
    Fy = q*crossy
    Fz = q*crossz

    xdd = Fx/m   # acceleration
    ydd = Fy/m
    zdd = Fz/m

    t = t + dt
    count = count + 1

##############################################

print dt
print t
print r

#>>> 
#1e-06
#0.00087
#2.53675834257
#>>> ================================ RESTART ================================
#>>> 
#1e-07
#0.000868
#2.53494066593
#>>> ================================ RESTART ================================
#>>> 
#1e-08
#0.000867870000002
#2.5347589641
#>>>
#1e-09
#0.000867852000009
#2.53474079446
#>>>