Magnetic Fields due to Currents
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
“Water, fire, air and dirt,
[freaking] magnets, how
do they work?”
- Insane Clown Posse
David J. Starling
Penn State Hazleton
PHYS 212
Magnetic Fields due to Currents
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Objectives
(a) Calculate the magnetic field due to a current using the
Biot-Savart Law.
(b) Relate the line integral of the magnetic field around a
closed path to the electric current enclosed by the path
(Ampere’s Law) and use this relationship to calculate
the magnetic field created by the current.
Solenoid and Toroid
Magnetic Field from a Current
Moving charges are affected by magnetic fields:
I ~
FB = q~v × ~B
I
But there is a symmetry to this force/field relationship
I
Moving charges also create magnetic fields
I
An empirical result:
µ0 i d~s × r̂
d~B =
(Biot-Savart Law)
4π r2
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Magnetic Field from a Current
d~B =
I
µ0 i d~s × r̂
4π r2
µ0 : permeability of free space or magnetic constant
10−6
kg/s2 A2
I
µ0 = 1.26 ×
I
r is distance between current and point of interest
I
r̂ points from current to point of interest
I
d~s points along current direction
m
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Magnetic Field from a Current
A long wire:
I
What is the field from a long wire with current i?
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Magnetic Field from a Current
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Let’s do the math:
I
I
I
Start with Biot-Savart law and fill in the gaps:
µ0 i ds sin(θ)
dB =
4π
r2
√
r = R 2 + s2
√
sin(θ) = sin(π − θ) = O/H = R/r = R/ R2 + s2
µ0
iR ds
dB =
4π (s2 + R2 )3/2
Magnetic Field from a Current
This field is from a small part of the wire:
µ0
iR ds
dB =
4π (s2 + R2 )3/2
But this is the total magnetic field:
Z
ds
µ0 iR s=∞
B =
2
4π s=−∞ (s + R2 )3/2
∞
s
µ0 i
=
4πR (s2 + R2 )1/2 −∞
µ0 i
=
[1 + 1]
4πR
µ0 i
=
2πR
The direction is given by the right-hand-rule.
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Magnetic Field from a Current
Chapter 29
Magnetic Fields due to
Currents
What about an arc of wire?
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
The total magnetic field:
Z
µ0
i ds sin(θ)
B =
4π
r2
ds = R dφ
sin(θ) = sin(π/2) = 1
R =
constant
i =
constant
Magnetic Field from a Current
Chapter 29
Magnetic Fields due to
Currents
What about an arc of wire?
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
The total magnetic field:
B =
=
=
Z
µ0 i φ R dφ sin(π/2)
4π 0
R2
Z φ
µ0 i
dφ
4πR 0
µ0 iφ
4πR
Magnetic Field from a Current
Question 1:
How does the result of the previous calculation change if we
include current in sections 1 and 2 in the figure below?
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
(a) The magnetic field is larger.
(b) The magnetic field is smaller.
(c) The magnetic field is the same.
(d) The magnetic field changes direction.
Magnetic Field from a Current
Example 1:
Find the magnetic field (magnitude and direction) at point P
from two parallel wires in the figure below; i1 = 15 A,
i2 = 32 A and d = 5.3 cm.
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Force Between Currents
How do two currents affect each other?
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
I
The first current creates a magnetic field:
µ0 ia
Ba =
2πd
I
The second current feels a force from this magnetic
field:
Fba = ib LBa =
µ0 Lia ib
2πd
Chapter 29
Magnetic Fields due to
Currents
Force Between Currents
How do two currents affect each other?
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Fba = ib LBa =
I
Parallel currents attract
I
Antiparallel currents repel
µ0 Lia ib
2πd
Force Between Currents
Question 2:
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Two parallel wires have currents that have the same
direction, but differing magnitude. The current in wire A is
i; and the current in wire B is 2i. Which one of the
following statements concerning this situation is true?
(a) Wire A attracts wire B with half the force that wire B attracts wire A.
(b) Wire A attracts wire B with twice the force that wire B attracts wire A.
(c) Both wires attract each other with the same amount of force.
(d) Wire A repels wire B with half the force that wire B attracts wire A.
(e) Wire A repels wire B with twice the force that wire B attracts wire A.
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Force Between Currents
Example 2:
Magnetic Field from a
Current
Force Between Currents
Four wires are in a square (a = 13.5 cm) as shown below,
each with a current of 7.50 A. Find the force per unit length
(N/m) on the bottom-left wire.
x
a
x
a
Ampere’s Law
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Ampere’s Law
Ampere’s Law
I
Magnetic Field from a
Current
Similar to Gauss’ Law, we can find the “enclosed
current”
Force Between Currents
Ampere’s Law
I
~B · d~s = µ0 ienc
(1)
Solenoid and Toroid
Ampere’s Law
Ampere’s Law
I
The right hand rule tells us if the current is positive or
negative
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
I
Although general, this law is often applied to problems
with symmetry
Ampere’s Law
Chapter 29
Magnetic Fields due to
Currents
Long straight wire:
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Let’s apply Ampere’s Law:
I
I
~B · d~s =
B cos(θ)ds
I
= B ds = B(2πr)
B(2πr) = µ0 ienc
µ0 i
B =
2πr
Ampere’s Law
Chapter 29
Magnetic Fields due to
Currents
Long thick wire:
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Let’s apply Ampere’s Law again:
I
I
~B · d~s =
B cos(θ)ds
I
= B ds = B(2πr)
B(2πr) = µ0 ienc
µ0 ienc
B =
2πr
Chapter 29
Magnetic Fields due to
Currents
Ampere’s Law
Long thick wire:
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
But what is ienc ?
I
For uniformly distributed current, we use a ratio:
ienc = i ×
I
πr2
= i(r/R)2
πR2
Therefore,
B=
µ0 i
r
2πR2
Force Between Currents
Example 3:
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Find the magnetic field at the dot in the current carrying
cylinder in the figure below. For this system, a = 2.0 cm,
b = 4.0 cm, r = 3.0 cm, and the current density (out of the
page) is given by J = cr2 where c = 3.0 × 106 A/m4 .
Ampere’s Law
Solenoid and Toroid
Solenoid and Toroid
Solenoid:
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Infinite Solenoid:
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
For this loop, there are four sections
I
Z b
Z c
Z d
Z a
~B · d~s =
~B · d~s +
~B · d~s +
~B · d~s +
~B · d~s
a
b
c
d
Z b
~B · d~s = Bh
=
a
= µ0 ienc
= µ0 Ni
= µ0 (nh)i
B = µ0 ni
(ideal solenoid)
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Question 3:
Magnetic Field from a
Current
When the switch below is closed, what happens to the
portion of the wire that runs inside of the solenoid?
Force Between Currents
Ampere’s Law
Solenoid and Toroid
(a) There is no effect on the wire.
(b) The wire is pushed downward.
(c) The wire is pushed upward.
(d) The wire is pushed toward the left.
(e) The wire is pushed toward the right.
Chapter 29
Magnetic Fields due to
Currents
Solenoid and Toroid
Toroid:
Magnetic Field from a
Current
Force Between Currents
Ampere’s Law
Solenoid and Toroid
This loop is just a simple circle
I
~B · d~s = B(2πr)
= µ0 ienc
= µ0 Ni
µ0 Ni 1
B =
2π r
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Magnetic Field from a
Current
Example 4:
Force Between Currents
Ampere’s Law
A solenoid of length 1.23 m and diameter 3.55 cm carries a
current of 5.57 A. There are five layers of loops, each with
850 turns along its length. What is the magnetic field at its
center?
Solenoid and Toroid
Solenoid and Toroid
Chapter 29
Magnetic Fields due to
Currents
Example 5:
Magnetic Field from a
Current
Find the magnetic field from a single loop of wire a distance
z away (shown below). The loop has radius R and current i.
Force Between Currents
Ampere’s Law
Solenoid and Toroid