SME
Electromagnetic Theory · Problem type
Ampère’s law: B inside and outside a thick wire
Use this for “A wire of radius 3 mm carries 30 A spread evenly over its cross-section. Find B 2 mm and 6 mm from its axis.”
Diagram
B
=
μ₀
Ir
2πa²
B
=
1.26 ×
10
-6
×
20
×
0.001
2π ×
0.002
2
B = 0.001 T
Inside the wire B grows with r; outside it falls as 1/r.
The peak is at r = a.
This is an example. Type your own values to start a new problem.
B (
)
×
2π
r (
)
=
μ₀
Ienc (
)
I
Current in the wire
example
a
Wire radius
example
Formulas
Ienc
=
I
×
(
r
a
)
2
5
=
20
×
(
0.001
0.002
)
2
B
×
2π
×
r
=
4π
×
10
-7
×
Ienc
0.001
×
2π
×
0.001
=
4π
×
10
-7
×
5
Formulas are worked in m, T. The step-by-step shows the conversions.
Assumptions
•
The wire is long, straight and round, radius a, with the current I spread evenly over its cross-section; μ₀ = 4π × 10⁻⁷ T·m/A.
•
By symmetry B circles the axis and is the same all round a circle of radius r, so Ampère’s law gives B × 2πr = μ₀Ienc.
•
Inside (r ≤ a) the circle takes in the share (r/a)² of the current; outside it takes in all of it.
Step-by-step
WE KNOW
I = 20 A, a = 2 mm, r = 1 mm
FIND
current inside r (Ienc), field at r (B)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to m, T).
a = 2 mm = 0.002 m (1 m = 1000 mm)
r = 1 mm = 0.001 m (1 m = 1000 mm)
Step 2 · Find current inside r (Ienc)
Use Ienc = I × (r ÷ a)²
Inside,
the
circle
takes
in
the
share
(
r
a
)
2
of
the
wire’s
cross-section,
so
that
share
of
I
.
0.001 ≤ 0.002: inside the wire
Ienc
=
20
×
(
0.001
0.002
)
2
Ienc
=
20
×
0.5
2
Ienc = 20 × 0.25
Ienc = 5 A
Step 3 · Find field at r (B)
Use B × 2π × r = 4π × 10⁻⁷ × Ienc
B
is
the
same
all
round
the
circle,
so
the
line
integral
is
B
×
2πr:
divide
μ₀Ienc
by
2πr.
B
=
4π
×
10
-7
×
Ienc
2π ×
r
B
=
4π
×
10
-7
×
5
2π × 0.001
B
=
0.001
T
Step 4 · Convert the answers
Change each answer back to the unit you chose.
B = 0.001 T = 1 mT (1 T = 1000 mT)
Check
Put the numbers back in. Both sides should match.
5
=
20
×
(
0.001
0.002
)
2
✓
0.001
×
2π
×
0.001
=
4π
×
10
-7
×
5
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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