SME
Electromagnetic Theory
Magnetostatics
Use this for “A loop of radius 8 cm carries 3 A. Find the field at its center and 6 cm along its axis.”
Diagram
B₀
=
μ₀
I
2R
=
1.26 ×
10
-6
×
5
2 × 0.1
B₀
=
3.14 ×
10
-5
T
B
=
μ₀IR²
÷
(2
(
z
2
+
R
2
)
3
2
)
=
1.11 ×
10
-5
T,
B
B₀
=
0.354
On the axis the sideways parts of dB from opposite pieces cancel.
This is an example. Type your own values to start a new problem.
B₀
=
μ₀
I (
)
2
×
R (
)
=
B₀ (
)
B
=
μ₀
(
)
2
2
(
(
z (
)
)
2
+
(
)
2
)
3
÷
2
=
B (
)
Formulas
B₀
=
4π
×
10
-7
×
I
2 ×
R
3.1416 ×
10
-5
=
4π
×
10
-7
×
5
2 × 0.1
B
=
4π
×
10
-7
×
I
×
R
2
÷
(2
×
(
z
2
+
R
2
)
3
2
)
1.1107 ×
10
-5
=
4π
×
10
-7
×
5
×
0.1
2
÷
(2
×
(
0.1
2
+
0.1
2
)
3
2
)
Formulas are worked in m, T. The step-by-step shows the conversions.
Assumptions
•
One thin circular loop in air; the point P is on its axis, z from the center, and μ₀ = 4π × 10⁻⁷ T·m/A.
•
Biot–Savart: each piece dl adds dB = μ₀I dl ÷ (4π(z² + R²)), square to the line from dl to P.
•
On the axis the sideways parts from opposite pieces cancel; only the parts along the axis add.
•
Far along the axis (z ≫ R) the loop acts as a magnetic dipole: B ≈ μ₀IR² ÷ (2z³).
Step-by-step
WE KNOW
I = 5 A, R = 10 cm, z = 10 cm
FIND
field at the center (B₀), field on the axis at z (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).
R = 10 cm = 0.1 m (1 m = 100 cm)
z = 10 cm = 0.1 m (1 m = 100 cm)
Step 2 · Find field at the center (B₀)
Use B₀ = 4π × 10⁻⁷ × I ÷ (2 × R)
At
the
center
every
piece
dl
is
R
away
and
square
to
the
line
to
it:
∮
μ₀
I
dl
4πR²
=
μ₀
I
2
R
.
B₀
=
4π
×
10
-7
×
5
2 × 0.1
B₀
=
3.1416 ×
10
-5
T
Step 3 · Find field on the axis at z (B)
Use B = 4π × 10⁻⁷ × I × R² ÷ (2 × (z² + R²)3/2)
Each
dl
is
√
z
2
+
R
2
from
P,
and
the
share
R
÷
√
z
2
+
R
2
of
its
dB
points
along
the
axis;
round
the
loop
that
adds
to
this.
B
=
4π
×
10
-7
×
5
×
0.1
2
÷
(2
×
(
0.1
2
+
0.1
2
)
3
2
)
B
=
1.1107 ×
10
-5
T
Step 4 · Convert the answers
Change each answer back to the unit you chose.
B₀ = 3.1416 × 10⁻⁵ T = 31.4159 μT (1 T = 10⁶ μT)
B = 1.1107 × 10⁻⁵ T = 11.1072 μT (1 T = 10⁶ μT)
Check
Put the numbers back in. Both sides should match.
3.1416 ×
10
-5
=
4π
×
10
-7
×
5
2 × 0.1
✓
1.1107 ×
10
-5
=
4π
×
10
-7
×
5
×
0.1
2
÷
(2
×
(
0.1
2
+
0.1
2
)
3
2
)
✓
← Previous lesson
The potential of an electric dipole, far away
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Ampère’s law: B inside and outside a thick wire
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