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Magnetic Field
The problem:
A square conducting loop with sides L and carrying current I is lying in the x − y plane, so that
its center is at the origin. Find the field on the z axis.
The solution:
The field of a finite wire at distance r is given by
~ =
dB
µ0 Id~l × (~r − ~r0 )
4π |~r − ~r0 |3
(1)
where ~r = (r, 0, 0), ~r0 = (0, 0, z 0 ) and d~l = (0, 0, dz 0 ) in the cylindrical coordinates (where we think
of the wire going into the ẑ 0 direction.) The field is in ϕ̂ direction (around each wire)
L
µ0 I
Bϕ =
r
4π
Z2
(r2
1
1
µ0 I L
dz 0 =
ϕ̂
3/2
02
2
4π r (r + (L/2)2 )1/2
+z )
(2)
−L
2
There are four wires that contribute to the field and we are interested in the ẑ component, so we
have to take four times the field we found and to consider the angle between ẑ and field’s direction
Bz = 4Bϕ cos θ = 4Bϕ
L
2r
(3)
where r2 = ( L2 )2 + z 2 .
For the projection onto the x − y plane see the Fig. below.
(on the Fig. the angle between r and B is π/2 because B is a tangent).
Finally
Bz =
µ0 2L2 I
1
q
4π ( L2 )2 + z 2 2( L )2 + z 2
(4)
2
1
Magnetic Field
The problem:
Two parallel half infinite wires are connected to a half ring wire with a radius R and carrying
current I. The half loop and the wires are in the x − y plane. What is the magnetic field in the
center of the half ring?
The solution:
This system can be duplicated and the duplicate system is put onto the original system, so that
one has two infinite wire and a ring between them and tangent to them. The field B created in the
center of the ring is twise the field created in the center of the half ring of the original system.
The field in the center of a ring is
B=
µ0 I
2r
(1)
The field of infinite wire is
B=
µ0 I
2πr
(2)
Both wires contribute in the same direction (ẑ)as the loop, therefore using superposition and dividing
by two, we get
µ0 I
µ0 I
µ0 I
1
2
+
=
(2 + π)
(3)
B=
2
2πr
2r
4πr
1
Magnetic Field
The problem:
In a ring with radius R an arc of an angle α has been replaced with a straight cord. In the ring
flows current I1 . A straight infinite wire passes through the center of the ring and carries a current
I2 . Calculate the force and the torque on the ring.
The solution:
Using the superposition principle we calculate the forces due to the infinite wire acting on the cord
and on the arc of an angle 2π − α and then sum them up.
The force acting on the arc
The magnetic field induced by the infinite wire is in the φ̂ direction and so is the element d~l = rdφφ̂.
Then the force and the torque are zero.
The force acting on the cord
Using the Biot - Savart’s law we can calculate the field on the cord
Z ~
µ0
dl × (~r − ~r0 )
~
B =
I
4π
|~r − ~r0 |3
(1)
Let the origin be at the center of the ring. The y-axis goes from the center of the ring toward the
cord and perpendicular to it.
So that we are interested in the field at
α
~r = (x, R cos , 0)
(2)
2
α
α
(3)
−R sin < x < +R sin
2
2
The position of the current element d~l is
~r0 = (0, 0, z)
d~l = dz ẑ
(4)
(5)
Then
α
~r − ~r0 = (x, R cos , −z)
2
r
α
~r − ~r0 =
x2 + R2 cos2 + z 2
2
α
d~l × (~r − ~r0 ) = dz(−R cos x̂ + xŷ)
2
(6)
(7)
(8)
Since the field in the x direction does not induce force, we calculate only the field in the y direction
Z ∞
xdz
2KI2 x
~
By = KI2
(9)
3 =
2
x + R2 cos2 α2
−∞ (x2 + R2 cos2 α + z 2 ) 2
2
1
where K =
µ0
4π .
Now we need to calculate the force and the torque:
~
= Id~l × B
d~τ = ~r × dF~
dF~
(10)
(11)
Then the force is
dF~
F~
= −I1 dxx̂ × By ŷ =
Z
=
dF~ = −
R sin
α
2
−R sin
α
2
Z
2KI1 I2 x
+ R2 cos2
dxẑ
(12)
I1 dxBy (x)ẑ = 0
(13)
x2
α
2
The integral is zero since the the integrand is an odd function.
The torque (relatively to the middle of the cord) is
2KI1 I2 x2
dxŷ
+ R2 cos2 α2
Z
Z R sin α
2
x2 dx
~τ =
d~τ = 2I1 I2 K ŷ
2
2
2
−R sin α x + R cos
d~τ = ~x × dF~ =
(14)
x2
2
2
α
2
α α
α
= 4I1 I2 KR sin − cos
ŷ
2
2
2
(15)