Physics 117
Group:_____ Names:__________________________________
Magnetic fields from currents
1. The drawing shows two long, thin wires that carry currents in the positive z direction. Both
wires are parallel to the z axis. The 50-A wire is in the x-z plane and is 5 m from the z axis. The
40-A wire is in the y-z plane and is 4 m from the z axis. What is the magnitude of the magnetic
field at the origin?
The magnetic field for a current carrying wire is
𝜇 𝐼
0
𝐵 = 2𝜋𝑟
Using this for the 40A and the 50 A
wires, we can determine the magnitude of the
magnetic field at the origin due to each of these
wires.
𝐵40𝐴 𝑥̂ =
𝜇0 (40𝐴)
𝑥̂
2𝜋(4𝑚)
𝐵50𝐴 (−𝑦̂) =
= 2𝜇𝑇𝑥̂ and
𝜇0 (50𝐴)
(−𝑦̂) = 2𝜇𝑇(−𝑦̂)
2𝜋(5𝑚)
Notice, the fields are not in the same direction so we have to use vector addition. The magnitude of the
magnetic field at the origin is therefore: 𝐵 = √(2𝜇𝑇)2 + (2𝜇𝑇)2 = 2.8𝜇𝑇
2. The figure shows a cross section of a long conducting coaxial cable with its
radii (a,b,c). Here a=0.40 cm, b=1.8 cm, and c=2.0 cm. Equal but opposite currents i
=5 A are uniformly distributed in the two conductors. Find the magnetic field B(r)
for various values of radial distance r. (a) r = 0.1 cm. (b) r = 1.6 cm. (c) r =1.9 cm. (d)
r=25 cm.
Ampere’s Law can be used to give the magnetic field in each of the regions of the coaxial cable.
⃗ ∙ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐
a) For r<a: B is constant for all space. ∫ 𝐵
𝜋𝑟 2
𝜋𝑎2
𝑟2
𝐵(2𝜋𝑟) = 𝜇0 𝐼 2
𝑎
𝜇0 𝐼𝑟
𝐵=
= 6.15𝑥10−5 𝑇
2𝜋𝑎2
b) For b>r>a: There is no extra current in the shell, just the smaller, enclosed wire. 𝐵 ∫ 𝑑𝑙 = 𝜇0 𝐼𝑒𝑛𝑐
𝐵 ∫ 𝑑𝑙 = 𝜇0 𝐼
𝐵(2𝜋𝑟) = 𝜇0 𝐼
𝜇0 𝐼
= 6.25𝑥10−5 𝑇
2𝜋𝑟
c) For the region b<r<c: The magnetic field from the inner wire will, by superposition, subtract
some of the magnetic field from the outer conductor (because the current is in opposite
𝐵=
𝜋𝑟 2 −𝜋𝑏2
directions). So, the ratio of the current density is:𝜋𝑐 2 −𝜋𝑏2
𝐵=
𝐵=
𝜇0 𝐼 𝜇0 𝐼 𝑟 2 − 𝑏 2
−
(
)
2𝜋𝑟 2𝜋𝑟 𝑐 2 − 𝑏 2
𝜇0 𝐼
𝑟2 − 𝑏2
(1 − 2
) = 2.7𝑥10−5 𝑇
2𝜋𝑟
𝑐 − 𝑏2
d) For the region r>c: The current enclosed is zero so the B field is 0. B=0.
3. Each of the following configurations has a square with side r and current I, in each wire. What is
the B field
in the
center of
each
square?
The B field from a wire is always 𝐵 =
𝜇0 𝐼
,
2𝜋𝑟
so what we have to do here, with the constant r’s
(from corner to center) is to consider directions.
Situation 1: Here the B fields from the upper left and the lower right will cancel and the B fields
from the lower left and the upper right will cancel. So B=0
Situation 2: In this case, the B fields all have components that will add together. 𝑑 =
𝑟√2
2
and
then there’s a cosine or sine factor but each corner is at 45 degrees, so we have the same √2⁄2
𝜇 𝐼 √2
)
2
0
factor for each term. The end result is that 𝐵 = 4 ( 𝑟√2
2
Situation 3: Again, the corners will cancel, so B=0
=
4𝜇0 𝐼
𝑟