SME
Electromagnetic Theory · Problem type
A coil in a uniform field: its moment, torque and energy
Use this for “A 120-turn coil of area 25 cm² carries 0.4 A in a 0.3 T field, its moment at 60° to the field. Find the torque and the energy.”
Diagram
μ = NIA = 50 × 0.5 × 0.01
μ
=
0.25
A·
m
2
τ = μB sin θ
τ = 0.25 × 0.2 × sin 30° = 0.025 N·m
U = −μB cos θ = −0.0433 J
The torque turns μ toward B.
This is an example. Type your own values to start a new problem.
μ
=
N
×
I (
)
A (
)
=
μ (A·m²)
τ
=
B (
)
sin
θ
°
=
τ (N·m)
U
=
−
B (
)
cos
θ
°
=
U (J)
Formulas
μ
=
N
×
I
×
A
0.25 = 50 × 0.5 × 0.01
τ
=
μ
×
B
×
sin(
θ
)
0.025 = 0.25 × 0.2 × sin(30°)
U
=
−
μ
×
B
×
cos(
θ
)
−0.0433 = −0.25 × 0.2 × cos(30°)
Assumptions
•
A flat coil of N turns, each of area A, carries I; its magnetic moment μ = NIA is square to its plane by the right-hand rule.
•
The field B is the same all over the coil, so the net force is 0 and only a torque τ = μB sin θ acts, turning μ toward B.
•
θ is the angle from μ to B; the energy U = −μB cos θ is least at θ = 0 and greatest at θ = 180°.
Step-by-step
WE KNOW
N = 50, I = 0.5 A, A = 0.01 m², B = 0.2 T, θ = 30°
FIND
magnetic moment (μ), torque (τ), energy in the field (U)
Step 1 · Find magnetic moment (μ)
Use μ = N × I × A
Each
turn
is
a
loop
of
moment
IA;
N
turns
side
by
side
add
N
of
them.
μ
=
50
×
0.5
×
0.01
μ
=
25
×
0.01
μ
=
0.25
A·
m
2
Step 2 · Find torque (τ)
Use τ = μ × B × sin(θ)
τ
=
μ
×
B
in
size:
largest
with
μ
square
to
B
,
0
with
μ
along
B
or
against
it.
τ
=
0.25
×
0.2
×
sin(30°)
τ
=
0.25
×
0.2
×
0.5
τ
=
0.05
×
0.5
τ
=
0.025
N·m
Step 3 · Find energy in the field (U)
Use U = −μ × B × cos(θ)
U
=
−
μ
·
B
:
the
work
the
field
would
do
turning
μ
back
to
θ
=
90°,
where
U
is
0.
U
=
−0.25
×
0.2
×
cos(30°)
U
=
−0.25
×
0.2
×
0.866
U
=
−0.05
×
0.866
U
=
−0.0433
J
Check
Put the numbers back in. Both sides should match.
0.25 = 50 × 0.5 × 0.01 ✓
0.025 = 0.25 × 0.2 × sin(30°) ✓
−0.0433 = −0.25 × 0.2 × cos(30°) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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