SME
University Physics II: Electricity & Magnetism · Problem type
A solenoid: its field, inductance and stored energy
Use this for “A 400-turn solenoid 20 cm long, with a 2 cm² cross-section, carries 1.5 A. Find the field inside, its inductance and the energy it stores.”
Diagram
n
=
N
ℓ
=
100
0.1
=
1,000
turns/m
B
=
μ₀nI
=
1.26 ×
10
-6
×
1,000
×
2
B = 0.00251 T
L
=
μ₀
N
2
A
ℓ
=
1.26 ×
10
-5
H
U = ½LI² = 2.51 × 10⁻⁵ J
Inside, the lines run straight and evenly spaced: B is the same everywhere.
This is an example. Type your own values to start a new problem.
B (
)
=
μ₀
×
I (
)
ℓ (
)
A
Cross-section area
example
n
Turn density
calculated · /m
L
Inductance
calculated
U
Energy stored in the field
calculated
Formulas
n
=
N
ℓ
1,000
=
100
0.1
B
=
4π
×
10
-7
×
n
×
I
0.002513
=
4π
×
10
-7
×
1,000
×
2
L
=
4π
×
10
-7
×
N
2
×
A
ℓ
1.2566 ×
10
-5
=
4π
×
10
-7
×
100
2
×
0.0001
0.1
U
=
1
2
×
L
×
I
2
2.5133 ×
10
-5
=
1
2
×
(1.2566
×
10
-5
)
×
2
2
Formulas are worked in m², T, J. The step-by-step shows the conversions.
Assumptions
•
The solenoid is long next to its width: inside, B runs along the axis and is the same everywhere; outside it is nearly 0.
•
Ampère’s law round a rectangle half inside the coil gives Bℓ = μ₀NI, so B = μ₀nI with n = N ÷ ℓ, and μ₀ = 4π × 10⁻⁷ T·m/A.
•
Each turn holds the flux BA, so NBA = LI gives L = μ₀N²A ÷ ℓ; the field stores U = ½LI².
•
The core is air or empty, so no iron adds to the field.
Step-by-step
WE KNOW
N = 100, ℓ = 0.1 m, A = 1 cm², I = 2 A
FIND
turn density (n), field inside (B), inductance (L), energy stored in the field (U)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m², T, J).
A = 1 cm² = 0.0001 m² (1 m² = 10⁴ cm²)
Step 2 · Find turn density (n)
Use n = N ÷ ℓ
The turns per meter: the turns spread over the length.
n
=
100
0.1
n
=
1,000
/m
Step 3 · Find field inside (B)
Use B = 4π × 10⁻⁷ × n × I
Ampère’s
law:
Bℓ
=
μ₀NI
round
a
rectangle
half
inside,
so
B
=
μ₀nI.
The
length
of
the
coil
cancels.
B
=
4π
×
10
-7
×
1,000
×
2
B
=
0.002513
T
Step 4 · Find inductance (L)
Use L = 4π × 10⁻⁷ × N² × A ÷ ℓ
N
turns
each
hold
the
flux
μ₀(
N
ℓ
)IA,
so
the
flux
linkage
per
amp
is
μ₀
N
2
A
ℓ
.
Doubling
N
gives
four
times
L
.
L
=
4π
×
10
-7
×
100
2
×
0.0001
0.1
L
=
1.25664 ×
10
-5
H
Step 5 · Find energy stored in the field (U)
Use U = ½ × L × I²
The
energy
the
field
holds:
the
work
done
against
the
back
emf
while
the
current
rose
from
0
to
I
.
U
=
1
2
×
(1.25664
×
10
-5
)
×
2
2
U
=
1
2
×
(1.25664
×
10
-5
)
×
4
U
=
(6.283
×
10
-6
)
×
4
U
=
2.5133 ×
10
-5
J
Step 6 · Convert the answers
Change each answer back to the unit you chose.
B = 0.002513 T = 2.5133 mT (1 T = 1000 mT)
U = 2.5133 × 10⁻⁵ J = 25.1327 μJ (1 J = 10⁶ μJ)
Check
Put the numbers back in. Both sides should match.
1,000
=
100
0.1
✓
0.002513
=
4π
×
10
-7
×
1,000
×
2
✓
1.25664 ×
10
-5
=
4π
×
10
-7
×
100
2
×
0.0001
0.1
✓
2.5133 ×
10
-5
=
1
2
×
(1.25664
×
10
-5
)
×
2
2
✓
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