SME
Grade 11 Science · Problem type
A charge circling in a magnetic field
Use this for “A proton moves at 3 × 10⁶ m/s square to a 0.2 T field. What force acts on it, what is the radius of its circle, and how long is one lap?”
Diagram
F = |q|vB
F
=
1.6 ×
10
-19
×
3 ×
10
6
×
0.2
F
=
9.61 ×
10
-14
N
r = mv/(|q|B)
r
=
1.67 ×
10
-27
×
3 ×
10
6
/(1.6
×
10
-19
×
0.2)
r = 0.157 m
The force is always square to v, so it turns the charge without speeding it up: a circle.
This is an example. Type your own values to start a new problem.
q
Charge
example · e
m
Mass
example
v
Speed
example · m/s
B
Magnetic field
example · T
F
Force
calculated
r
Radius
calculated · m
T
Time for one lap
calculated · s
Formulas
F
=
q
×
1.602 ×
10
-19
×
v
×
B
9.61 ×
10
-14
=
1
×
1.602 ×
10
-19
×
(3
×
10
6
)
×
0.2
r
=
m
×
v
/(
q
×
1.602 ×
10
-19
×
B
)
0.157
=
1.673 ×
10
-27
×
(3
×
10
6
)/(1
×
1.602 ×
10
-19
×
0.2)
T
=
2π
×
r
v
3.28 ×
10
-7
=
2π
×
0.157/(3
×
10
6
)
Assumptions
•
The field is square to the velocity, so F = |q|vB, with q counted in charges of e = 1.602 × 10⁻¹⁹ C.
•
The force is always square to v: it turns the charge without speeding it up, so the path is a circle.
•
That force is the centripetal force: |q|vB = mv²/r, so r = mv/(|q|B).
•
A proton’s mass is 1.673 × 10⁻²⁷ kg; an electron’s is 9.109 × 10⁻³¹ kg.
Step-by-step
WE KNOW
q = 1 e, m = 1.673 × 10⁻²⁷ kg, v = 3 × 10⁶ m/s, B = 0.2 T
FIND
force (F), radius (r), time for one lap (T)
Step 1 · Find force (F)
Use F = q × 1.602 × 10⁻¹⁹ × v × B
The charge in coulombs times the speed and the field.
F
=
1
×
1.602 ×
10
-19
×
(3
×
10
6
)
×
0.2
F
=
(4.806
×
10
-13
)
×
0.2
F
=
9.61 ×
10
-14
N
Step 2 · Find radius (r)
Use r = m × v/(q × 1.602 × 10⁻¹⁹ × B)
Momentum over charge times field: a faster or heavier charge circles wider.
r
=
1.673 ×
10
-27
×
(3
×
10
6
)/(1
×
1.602 ×
10
-19
×
0.2)
r
=
(1.673
×
10
-27
)
×
(3
×
10
6
)/(3.204
×
10
-20
)
r
=
(5.019
×
10
-21
)/(3.204
×
10
-20
)
r
=
0.1566
m
Step 3 · Find time for one lap (T)
Use T = 2π × r/v
One
lap
of
the
circle
at
speed
v
.
A
faster
charge
circles
wider,
so
the
lap
takes
the
same
time.
T
=
2π
×
0.1566/(3
×
10
6
)
T
=
3.28 ×
10
-7
s
Check
Put the numbers back in. Both sides should match.
9.61 ×
10
-14
≈
1
×
1.602 ×
10
-19
×
(3
×
10
6
)
×
0.2
✓
0.1566
≈
1.673 ×
10
-27
×
(3
×
10
6
)/(1
×
1.602 ×
10
-19
×
0.2)
✓
3.28 ×
10
-7
≈
2π
×
0.1566/(3
×
10
6
)
✓
Topic check
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