Download P132 Chapter 30

Chapter 30: Magnetic Fields Due to Currents
A moving electric charge creates a magnetic field. One of the more practical ways of
generating a large magnetic field (0.1-10 T) is to use a large current flowing through a wire.
The calculation of the magnetic field produced by a wire can be very complicated as it
involves vector multiplication and integration! In the early 1800’s it was realized by Biot and
Savart that the magnetic field due to a conductor carrying a current could be expressed as:
dB =
ds
µ 0 Ids × r
4π r 3
The variables are defined in the picture.
In this equation µ0 is the permeability constant:
µ0=4πx10-7 T-m/A
θ
r
dB
P
R
R. Kass
B is into the page
The magnitude of B is given by:
µ Ids sin θ
dB = 0
4π
r2
θ is the angle
between r and ds
Some things to keep in mind about the Biot Savart Law:
a) The direction of dB is perpendicular to the vectors ds and r.
b) The magnitude of dB varies as 1/r2.
c) The magnitude of dB is proportional to the current (I) and
the length of the current element, ds,
d) The magnitude of dB is proportional to the sine of the angle
between the vectors ds and and r.
e) To get the B field from a finite size wire we must integrate dB!
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Direction of Magnetic Field due to Current in a Wire
The second Right Hand rule
The direction of the magnetic field due to a current in a wire can be found using your right hand:
1) Your thumb points in the direction of the current
2) The direction that your fingers “naturally” curl gives the direction of the magnetic field.
●=B out of the page
x=B into the page
HRW 30-2
Example showing the lines of the magnetic field for a current
going into the page. The field lines form concentric circles around
the wire. The direction of the B-field at a point is tangent to the
circle at the point of interest. In this figure a vector representing
the magnetic field is drawn for two different distance from the wire.
The relative size of the vectors gives the relative magnitudes of the
B-field at the two point.
Another view of the relationship between the
direction of the current and the magnetic field.
HRW 30-4
R. Kass
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Force on a Wire from Another Wire
Question:
Two wires are next to each other as in the figure. For both wires the current is INTO the page.
What is the direction of the magnetic force on wire 2 due to the magnetic field of wire 1?
First we need to find the direction of: L × B
HRW 30-2
FB
wire 2
X
wire 1
B
HRW 30-2
B
FB
What about the force on wire 1 due to wire 2?
The magnetic field at wire 1 due to wire 2 is pointed
towards the top of the page. Using the (1st) right hand
rule (cross products) we find that
wire 1 is attracted to wire 2.
X
wire 1
R. Kass
The direction of L follows the current (positive charge flow)
and is into the paper. The direction of B at wire 2 is found
using the 2nd right hand rule and is pointed DOWN. The
direction of the vector cross product is towards wire 1.
The force is also in this direction since we are assuming the
current is made of positive charge carriers.
Therefore wire 2 is attracted to wire 1.
wire 2
If the currents in the wires run in opposite directions
then the wires will repel each other.
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Calculation of the Magnetic Field Due to a Current in a wire
Let’s consider the simple case (which is not so simple) of a long (length =L) wire carrying
a constant current (I). We want to calculate B at a point along the middle of the wire a distance
L/2
R away from the wire.
µ
Ids × r
B=
0
4π
∫
−L / 2
r3
Since the current is constant we can pull it outside of the integral:
L/2
µ0 I
ds × r
B=
4π
r3
∫
−L / 2
ds
s
For this situation dB is always into the page, no matter where we are along the wire.
So, there is only one component of B. This simplifies things and we now have:
θ
π−θ
µ I
B= 0
4π
r
dB
P
L
R
B is into the page
L/2
∫
ds sin θ
−L / 2
r2
no longer a vector equation!
We now need to do a bit of trig:
r = s2 + R2
µ I
B= 0
4π
B=
L/2
∫
−L / 2
sin θ = sin(π − θ ) = R / r = R / s 2 + R 2
Rds
(s 2 + R 2 )3 / 2

µ0 I 
L


4πR  ( L2 / 4 + R 2 )1/ 2 
For a very long wire, L>>R, this simplifies to: B =
R. Kass
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L/2

µ IR 
s
Integral #19
= 0  2 2

4π  R ( s + R 2 )1/ 2  − L / 2 Appendix E
If L2>>R2 then (L2/4+R2)1/2=L/2
µ0 I
2πR
The force between two wires
separated by a distance d is:
µ II L
F= 0 1 2
2πd
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Ampere’s Law
When we studied electrostatics we found that for certain problems with a high degree of
symmetry (e.g. cylinder, sphere, plane) we could find the electric field easier using Gauss’s Law
instead of direct integration.
∫ E ⋅ dA = q
enc
/ε0
Our old friend, Gauss’s Law
A similar situation exists when the magnetic field is generated by a current.
∫ B ⋅ ds = µ I
Our new friend, Ampere’s Law
0 enc
Like Gauss’s Law, Ampere’s law contains an integral with a dot product of vectors. It also
involves an integral over a specific path.
For Gauss’s Law the integration was over a closed surface that enclosed charge.
For Ampere’s Law the integration is over a closed loop that encloses current.
The right side of Ampere’s Law includes all currents enclosed by the loop .
Let’s do some examples with current carrying wires to illustrate the features of Ampere’s Law.
A wire with a dot has current coming out of the page, one with an x has current into the page.
Ampereian
Ampereian
loop
I2
loop
I
2
I1
I1 X
X
I
1
●
I1
●
●
●
The current enclosed by the
Amperian loop is I1 for both
examples.
R. Kass
I3
●
I1
●
I2
X
I5
I4
X
●
The current enclosed by these
Amperian loop is I1–I2.
If the currents are equal but the
directions opposite then Ienc=0.
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The current enclosed by this
Amperian loop is still I1–I2.
Currents outside the loop
do not contribute to Ienc.
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B ⋅ ds = µ I
∫
We now need a sign convention for the enclosed current. We use the convention of the text to
Ampere’s Lawcontinued
0 enc
define the direction of integration using the fingers of the right hand for the loop direction and
the thumb to assign the direction of positive enclosed current.
HRW Fig 30-12
Following our convention, Ienc=I1-I2.
Since the integral contains the dot product between ds and B we need to define the angle between
these vectors.
The vector ds is always tangent to the curve and is oriented along the
direction of integration, counter clockwise according to our convention.
The angle θ is the angle between ds and B as shown in the figure.
Putting it all together we have for this example:
HRW Fig 30-11
∫
∫ B cosθds = µ ( I
B ⋅ ds = µ 0 I enc
0
1
− I2 )
The current I3 does not contribute to Ienc
since it is outside the loop.
BUT this current does contribute to B!
Without more information we can not solve for B.
All we know is that the integral is equal to µ0(I1-I2).
R. Kass
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Applications of Ampere’s Law
Just like Gauss’s Law there are only a few instances where we can actually solve for B.
One such case is the magnetic field outside a long straight wire.
HRW Fig.30-13
For a long straight wire the magnetic field is only a function of the
distance from the wire. We want to exploit this piece of info
to pick our Amperian loop. We pick a circle with the wire at the center
for our Amperian loop because B has the same value at every point on
the circle. In addition (extra bonus), since ds is tangent to the circle it
is parallel to B everywhere so cosθ=1. The integral is evaluated around
the circle (remember arc length, ds =rθ)
2π
2π
0
0
∫ B ⋅ ds = ∫ B(cos 0)ds = ∫ Bds = B ∫ rdθ = Br ∫ dθ = 2πrB
Ampere’s Law says: ∫ B ⋅ ds = µ 0 I enc
2πrB = µ 0 I
B=
µ0 I
2πr
Same result as from direct integration!!
Example: B inside a wire (of radius R) with uniform current (I).
Since we have a uniform current inside the wire Ienc=I(πr2)/(πR2). Since we still have a long straight wire
the integral part of the problem is identical to the one we just solved.
2πrB = µ 0 I enc
R. Kass
µ0 I
πr 2
= µ0 I
⇒
=
(
)r
B
2
2
πR
πR
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Applications of Ampere’s Lawcontinued
Another situation where Ampere’s Law is useful is to calculate the B-field inside a solenoid.
A solenoid is just a hollow cylinder with wire wrapped around it as shown in the figure.
B
We can use Ampere’s law to find B inside the
solenoid by breaking the integral into 4 pieces.
HRW Fig. 30-16
b
c
a
b
In an ideal solenoid there is a uniform magnetic
field inside the coils and zero magnetic field outside.
For current flowing in the direction shown in the figure
the B-field points to the right.
d
a
c
d
our loop
∫ B ⋅ ds = ∫ B ⋅ ds + ∫ B ⋅ ds + ∫ B ⋅ ds + ∫ B ⋅ ds
The integrals from b to c and d to a are zero since
ds is perpendicular to B (cosθ=0). The integral c to d
is zero since B=0 in this region. So we are left with:
b
b
a
a
∫ B ⋅ ds = ∫ B ⋅ ds = B ∫ ds = Bh
HRW Fig. 30-19
The enclosed current is just the current in a turn (I) times the number of turns per length (n)
times the length (h): Ienc=nhI. So, B for an ideal solenoid is:
R. Kass
B=µ0nI
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