Cycloid Motion with EM Fields

## **Problem**

This is a fascinating electromagnetism problem I found in *Introduction to Electrodynamics* by David J. Griffifths (pages 213 - 216 in the 4th edition). The author did not explicitly solve the entire problem. Here I present my full solution.

Consider a uniform magnetic field directed in the x-direction, and a uniform electric field directed in the z-direction. A particle with positive charge

q

{\displaystyle q}

 and mass

m

{\displaystyle m}

 is released from the origin, and initially at rest. Determine the trajectory of the particle over time.

# E

(

0

0

E

)

,

# B

(

B

0

0

)

.

{\displaystyle \boldsymbol{E} =
\begin{pmatrix} 0 \\ 0 \\ E \end{pmatrix}\ , \quad \boldsymbol{B} =
\begin{pmatrix} B \\ 0 \\ 0 \end{pmatrix}\ . }

## **Solution**

The magnetic force is calculated using the cross-product

v
×
B

=

i

j

k

0

y
˙

z
˙

B

0

0

=

(

0

B

z
˙

−
B

y
˙

)

.

{\displaystyle \boldsymbol{v \times B} = \begin{vmatrix} \boldsymbol{i}&\boldsymbol{j}&\boldsymbol{k} \\
 0 & \dot y & \dot z \\
 B & 0 & 0 \end{vmatrix}\ =
\begin{pmatrix} 0 \\ B \dot z \\ -B \dot y \end{pmatrix}\ . }

Using the above cross-product, write the Lorentz force

# F
q

(

E

v
×
B

)

{\displaystyle \boldsymbol{F} = q \left(\boldsymbol{E} + \boldsymbol{v \times B} \right) }

as a three-dimensional vector

(

m

x
¨

m

y
¨

m

z
¨

)

=

(

0

q
B

z
˙

q
E
−
q
B

y
˙

)

.

{\displaystyle \begin{pmatrix} m \ddot x \\ m \ddot y \\ m \ddot z \end{pmatrix}\ =
\begin{pmatrix} 0 \\ qB \dot z \\ qE - qB \dot y \end{pmatrix}\ . }

Divide the mass of the particle on both sides of the equation. Let

# ω

q
B

m

{\displaystyle \omega = \frac{qB}{m} }

 and

# γ

q
E

m

{\displaystyle \gamma = \frac{qE}{m} }

, then

(

x
¨

y
¨

z
¨

)

=

(

0

ω

z
˙

γ
−
ω

y
˙

)

.

{\displaystyle \begin{pmatrix} \ddot x \\ \ddot y \\ \ddot z \end{pmatrix}\ =
\begin{pmatrix} 0 \\ \omega \dot z \\ \gamma - \omega \dot y \end{pmatrix}\ . }

To solve this system of coupled differential equations, let

x
¨

=

v
˙

x

,

y
¨

=

v
˙

y

,

z
¨

=

v
˙

z

{\displaystyle \ddot x = \dot v\_x, \quad \ddot y = \dot v\_y, \quad \ddot z = \dot v\_z }

. Thus

(

v
˙

x

v
˙

y

v
˙

z

)

=

(

0

ω

v
z

γ
−
ω

v
y

)

.

{\displaystyle \begin{pmatrix} \dot v\_x \\ \dot v\_y \\ \dot v\_z \end{pmatrix}\ =
\begin{pmatrix} 0 \\ \omega v\_z \\ \gamma - \omega v\_y \end{pmatrix}\ . }

Substitute the middle equation, into the bottom equation to obtain

v
¨

y

ω

2

v
y

=
ω
γ
.

{\displaystyle \ddot v\_y + {\omega}^{2} v\_y = \omega \gamma . }

The above differential equation is non-homogeneous, thus the general solution is the sum of the homogeneous solution and particular solution.

**(1) Homogeneous solution**

In order to solve the homogeneous solution, consider the differential equation

v
¨

h

ω

2

v
h

=
0

{\displaystyle \ddot v\_h + {\omega}^{2} v\_h = 0 }

and note that the characteristic equation is

λ

2

ω

2

=
0

{\displaystyle {\lambda}^{2} + {\omega}^{2} = 0 }

, where

# λ
±
ω
i

{\displaystyle \lambda = \pm \omega i }

. The solution is therefore

v
h

(
t
# )
A

e

i
ω
t

B

e

−
i
ω
t

{\displaystyle v\_h (t) = A{e}^{i \omega t} + B{e}^{-i \omega t} }

or

v
h

(
t
# )
C
sin
⁡

(
ω
t
)

D
cos
⁡

(
ω
t
)

.

{\displaystyle v\_h (t) = C\sin{(\omega t)} + D\cos{(\omega t)} . }

**(2) Particular solution**

Since

ω
γ

{\displaystyle \omega \gamma }

 is constant, if we let

v
p

=
ω
γ

{\displaystyle v\_p = \omega \gamma }

, then

v
˙

p

=

v
¨

p

=
0

{\displaystyle \dot v\_p = \ddot v\_p = 0 }

. Then the differential equation

v
¨

p

ω

2

v
p

=
ω
γ

{\displaystyle \ddot v\_p + {\omega}^{2} v\_p = \omega \gamma }

reduces to

0

ω

2

v
p

=
ω
γ
.

{\displaystyle 0 + {\omega}^{2} v\_p = \omega \gamma . }

Hence, we discover that

v
p

=

γ
ω

{\displaystyle v\_p = \frac{\gamma}{\omega}}

. Consequently, the solution of the differential equation is

v
y

(
t
# )
C
sin
⁡

(
ω
t
)

D
cos
⁡

(
ω
t
)

γ
ω

.

{\displaystyle v\_y (t) = C\sin{(\omega t)} + D\cos{(\omega t)} + \frac{\gamma}{\omega}. }

Since the particle is initially released at rest, the velocities in the y-direction and z-direction are

v
y

(
t
)

=

γ
ω

(

1
−
cos
⁡

(
ω
t
)

)

v
z

(
t
)

=

γ
ω

sin
⁡

(
ω
t
)

{\displaystyle \begin{align}
v\_y (t) &= \frac{\gamma}{\omega}\left(1 - \cos{(\omega t)}\right)\\[5 pt]
v\_z (t) &= \frac{\gamma}{\omega}\sin{(\omega t)}
\end{align}}

To obtain the trajectory, integrate each component velocity

∫

v
y

d
t

=

γ
ω

t
−

γ

ω

2

sin
⁡

(
ω
t
)

y
0

∫

v
z

d
t

=
−

γ

ω

2

cos
⁡

(
ω
t
)

z
0

{\displaystyle
\begin{align}
\int v\_y \; dt &= \frac{\gamma}{\omega}t - \frac{\gamma}{{\omega}^{2}}\sin{(\omega t)} + y\_0 \\[5 pt]
\int v\_z \; dt &= - \frac{\gamma}{{\omega}^{2}}\cos{(\omega t)} + z\_0
\end{align} }

Since the particle is initially released at the origin,

y
(
t
)

=

γ
ω

t
−

γ

ω

2

sin
⁡

(
ω
t
)

z
(
t
)

=

γ

ω

2

(

1
−
cos
⁡

(
ω
t
)

)

{\displaystyle
\begin{align}
y(t) &= \frac{\gamma}{\omega}t - \frac{\gamma}{{\omega}^{2}}\sin{(\omega t)}\\[5 pt]
z(t) &= \frac{\gamma}{{\omega}^{2}} \left(1 - \cos{(\omega t)}\right)
\end{align} }

Replacing

γ

{\displaystyle \gamma}

 and

ω

{\displaystyle \omega}

 with its original quantities

y
(
t
)

=

E
B

[

t
−

m

q
B

sin
⁡

(

q
B

m

t

)

]

z
(
t
)

=

E
m

q

B

2

[

1
−
cos
⁡

(

q
B

m

t

)

]

.

{\displaystyle
\begin{align}
y(t) &= \frac{E}{B}\left[t - \frac{m}{qB}\sin{\left(\frac{qB}{m} t\right)}\right]\\[5 pt]
z(t) &= \frac{Em}{q{B}^{2}}\left[1 - \cos{\left(\frac{qB}{m} t \right)}\right].
\end{align} }

The solution of the trajectory is a cycloid!