SME
University Physics II: Electricity & Magnetism · Problem type
A charged particle circling in a magnetic field
Use this for “An alpha particle (charge 2e, mass 6.645 × 10⁻²⁷ kg) moves at 1.5 × 10⁶ m/s square to a 0.8 T field. Find the radius of its circle and its cyclotron frequency.”
Diagram
F = |q|vB
F
=
1.6 ×
10
-19
×
2 ×
10
6
×
0.5
F
=
1.6 ×
10
-13
N
r = mv/(|q|B)
r
=
1.67 ×
10
-27
×
2 ×
10
6
/(1.6
×
10
-19
×
0.5)
r = 0.0418 m
The force is always square to v, so it turns the charge without speeding it up: a circle.
Cyclotron frequency f = 7.62 MHz
This is an example. Type your own values to start a new problem.
r (
)
=
×
|
|e
×
B (
)
f
Cyclotron frequency
calculated
Formulas
r
=
m
×
v
|
q
| × 1.602 ×
10
-19
×
B
0.04177
=
1.673 ×
10
-27
×
2 ×
10
6
|1| × 1.602 ×
10
-19
× 0.5
f
=
|
q
|
×
1.602 ×
10
-19
×
B
2π ×
m
7,620,030
=
|1|
×
1.602 ×
10
-19
×
0.5
2π × (1.673 ×
10
-27
)
Formulas are worked in m, Hz. The step-by-step shows the conversions.
Assumptions
•
The velocity is square to an even field and well below light speed, so the force |q|vB turns the charge without speeding it up: the path is a circle.
•
That force is the centripetal force, |q|vB = mv² ÷ r, so r = mv ÷ (|q|B). q is counted in charges of e = 1.602 × 10⁻¹⁹ C; its sign sets the way round.
•
One lap takes 2πr ÷ v, so the laps per second are f = |q|B ÷ (2πm): the cyclotron frequency is the same at every speed.
•
Electron: q = −1, m = 9.109 × 10⁻³¹ kg. Proton: q = 1, m = 1.673 × 10⁻²⁷ kg. Alpha: q = 2, m = 6.645 × 10⁻²⁷ kg.
Step-by-step
WE KNOW
q = +1 e, m = 1.673 × 10⁻²⁷ kg, v = 2 × 10⁶ m/s, B = 0.5 T
FIND
radius of the circle (r), cyclotron frequency (f)
Step 1 · Find radius of the circle (r)
Use r = m × v ÷ (|q| × 1.602 × 10⁻¹⁹ × B)
Set
|
q
|vB
=
mv
²
r
and
solve
for
r
:
momentum
over
charge
times
field,
so
heavier
or
faster
charges
circle
wider.
r
=
1.673 ×
10
-27
×
2 ×
10
6
|1| × 1.602 ×
10
-19
× 0.5
r
=
(1.673
×
10
-27
)
×
2 ×
10
6
(1.602 ×
10
-19
) × 0.5
r
=
(1.673
×
10
-27
)
×
2 ×
10
6
8.01 ×
10
-20
r
=
3.346 ×
10
-21
8.01 ×
10
-20
r
=
0.04177
m
Step 2 · Find cyclotron frequency (f)
Use f = |q| × 1.602 × 10⁻¹⁹ × B ÷ (2π × m)
Laps
per
second
are
v
2πr
;
with
r
=
mv
|
q
|
B
the
speed
cancels.
f
=
|1|
×
1.602 ×
10
-19
×
0.5
2π × (1.673 ×
10
-27
)
f
=
7,620,030.4493
Hz
Step 3 · Convert the answers
Change each answer back to the unit you chose.
r = 0.04177 m = 4.1773 cm (1 m = 100 cm)
f = 7,620,030.4493 Hz = 7.62 MHz (1 MHz = 10⁶ Hz)
Check
Put the numbers back in. Both sides should match.
0.04177
=
1.673 ×
10
-27
×
2 ×
10
6
|1| × 1.602 ×
10
-19
× 0.5
✓
7,620,030
=
|1|
×
1.602 ×
10
-19
×
0.5
2π × (1.673 ×
10
-27
)
✓
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