A point charge q with mass m is released from rest at the origin in an external magnetic field and an external electric
field. Both of the external fields are uniform but the electric field points in the z-direction and the magnetic field points
in the x-direction:
E~ = E0
ˆk B~ = B0ˆı
where E0 and B0 are constants. The electric field accelerates the point charge along the z-axis, but once the point charge
has acquired a non-zero velocity along the z-direction it will also experience a magnetic force.
a) Using Newton’s second law and the Lorentz force, show that the acceleration of the point charge in each direction must
satisfy the following:
d2x
dt2 = 0
d2y
dt2
= ω
dz
dt
d2z
dt2
= ωE0
B0
−
dy
dt
where ω is the cyclotron frequency
ω =
qB0
m
and x(t), y(t), and z(t) give the position of the point charge along each axis as functions of time.
b) The above equations are called coupled differential equations because they mix derivatives of y with derivatives of z
(and vice versa). Show that these equations can be de-coupled by taking another time derivative. By de-coupled, I mean
you should have an equation containing only derivatives of y with respect to t and an equation containing only derivatives
of z with respect to t.
c) The de-coupled equations are easier to solve, but let’s not go through the actual steps. Instead I will just give you the
general solution to the equations given in part a):
x = C1t + C2
y = C3 cos ωt + C4 sin ωt +
E0t
B0
+ C5
z = C4 cos ωt − C3 sin ωt + C6
where C1, C2, C3, C4, C5, and C6 are arbitrary constants. Show that these functions solve the equations from part a).
You do NOT need to derive them, just plug them directly into the equations from part a).
d) Recall that at t = 0 the point charge is at rest and located at (x, y, z) = (0, 0, 0). Use these initial conditions to
determine the values of the arbitrary constants.
e) (Optional) Now try to sketch the motion of the charged particle. This can be tricky. It is helpful to take your expressions
for y(t) and z(t) and notice that you can manipulate them into the equation for a circle, but one in which the center of
the circle moves along the y-axis with constant speed. This is called cycloid motion and is identical to a wheel rolling at
constant velocity
- A Point Charge Q With Mass M Is Released From Rest At The Origin In An External Magnetic Field And An External Electric 1 (63.32 KiB) Viewed 10 times
B = Boi 1. A point charge q with mass m is released from rest at the origin in an external magnetic field and an external electric field. Both of the external fields are uniform but the electric field points in the 2-direction and the magnetic field points in the e-direction: E = Ek where Eo and Be are constants. The electric field accelerates the point charge along the z-axis, but once the point charge has acquired a non-zero velocity along the s-direction it will also experience a magnetic force. a) Using Newton's second law and the Lorentz force, show that the acceleration of the point charge in each direction must satisfy the following: Edy d12 dla B dt where w is the cyclotron frequency qB. fy 0 w dt m and 2(0), y(t), and 30 give the position of the point charge along each axis as functions of time. b) The above equations are called coupled differential equations because they mix derivatives of y with derivatives of: (and vice versa). Show that these equations can be de-coupled by taking another time derivative. By de-coupled, I mean you should have an equation containing only derivatives of y with respect to t and an equation containing only derivatives of : with respect to t. c) The de-coupled equations are ensier to solve, but let's not go through the actual steps. Instead I will just give you the general solution to the equations given in part a): r = {+C y = Cg cost + Casinw + Eot + Cs Bo * = Cocoswt - Cssin wt+C6 where C1, C2, C3, C, Cs, and Co are arbitrary constants. Show that these functions solve the equations from part a). You do NOT need to derive them, just plug them directly into the equations from part a). d) Recall that at t = 0 the point charge is at rest and located at (7.7) = (0,0,0). Use these initial conditions to determine the values of the arbitrary constants. e) (Optional) Now try to sketch the motion of the charged particle. This can be tricky. It is helpful to take your expressions for y(t) and (t) and notice that you can manipulate them into the equation for a circle, but one in which the center of the circle moves along the y-axis with constant speed. This is called cycloid motion and is identical to a wheel rolling at constant velocity