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PHYSICS 1B
Today’s lecture:
The Biot-Savart Law
and
Ampere’s Law
Electricity & Magnetism
22/08/2013
PHYSICS 1B – Sources of the Magnetic Field
The Biot-Savart Law - Introduction
Jean-Baptiste Biot and Félix Savart conducted experiments on the
force exerted by an electric current on a nearby magnet.
They arrived at a mathematical expression that gives the magnetic
field at some point in space due to a current.
The magnetic field described by the Biot-Savert Law is the field due
to a given current carrying conductor.
Be careful not to confuse this field with any external field applied to the
conductor from some other source!
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
The Biot-Savart Law – Set-Up
Biot and Savart measured the magnetic field, dB, at
some distance r from a wire, which was carrying a
steady current of I.
The length element for a small section of the wire
is ds.
They found that dB is:
• proportional to the current, I, the length, ds and sin θ
• inversely proportional to r2
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
The Biot-Savart Law – Set-Up
They found that dB is proportional to the current, I, the length,
ds and sin θ, and inversely proportional to r2
These observations are brought together in the
Biot-Savart law:
𝜇0 𝐼 𝑑𝒔 × 𝒓
𝑑𝑩 =
4𝜋
𝑟2
This introduces a new constant, μ0, which is called the permeability of free
space.
T. m
−7
𝜇0 = 4𝜋 × 10
A
The magnetic field is due to the current-carrying conductor.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Total Magnetic Field
dB is the field created by the current in the length segment ds.
To find the total field, B, we have to sum up all the contributions from all the
current elements I ds .
Therefore:
𝜇0
𝑩=
4𝜋
𝐼 𝑑𝒔 × 𝒓
𝑟2
which is the integral over the entire current distribution.
The Biot-Savart Law is also valid for a current consisting of charges flowing
through space (i.e. we could apply it to the particle beam in an accelerator).
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Quick Quiz
Consider the current in the length of wire shown below.
Rank the points A, B, and C in terms of the magnitude of the magnetic field due
to the current in the length element, ds, shown, from the greatest to the least.
a) A, B, C
b) B, C, A
c) C, B, A
d) C, A, B
e) An equal field applies at all these points
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Quick Answer
Consider the current in the length of wire shown below.
Rank the points A, B, and C in terms of the magnitude of the magnetic field due
to the current in the length element, ds, shown, from the greatest to the least.
b) B, C, A
Point B is the closest to the current element. Point C is farther away, and in
addition, the field is reduced by the sin θ factor in the cross product 𝑑𝒔 × 𝒓.
The field at A is zero because θ = 0.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field Compared to Electric Field
Distance
The magnitude of the magnetic field varies as the inverse square of the distance
from the source.
The electric field due to a point charge also varies as the inverse square of the
distance from the charge.
Direction
The electric field created by a point charge is radial in direction.
The magnetic field created by a current element is perpendicular to both the
length element ds and the unit vector 𝒓.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field Compared to Electric Field
Source
An electric field is established by an isolated electric charge.
The current element that produces a magnetic field must be a part of an
extended current distribution, and so you therefore have to integrate over the
entire distribution of current to work out the magnetic field.
End of Field Lines
Magnetic field lines have no beginning and no end – they instead form
continuous circles.
Electric field lines begin on positive charges and end on negative charges.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field for a Long, Straight Conductor
First, find the field contribution from a small
element of the current, then integrate over
the current distribution.
We have thin, straight wire carrying a
constant current, I, so therefore:
𝑑𝒔 × 𝒓 = 𝑑𝑥 cos 𝜃 𝒌
So now, we just integrate over the current
elements from one end of the wire to the
other…
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field for a Long, Straight Conductor
So now, we just integrate over the current
elements from one end of the wire to the
other…
𝜇0 𝐼
𝑩=−
4𝜋𝑎
∴
𝜃2
cos 𝜃 𝑑𝜃
𝜃1
𝜇0 𝐼
𝑩=
sin 𝜃1 − sin 𝜃2
4𝜋𝑎
This is example 30.1 in the text book…
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field for a Long, Straight Conductor
Special Case
𝜇0 𝐼
𝑩=
sin 𝜃1 − sin 𝜃2
4𝜋𝑎
If the conductor is an infinitely long, straight
wire, then 𝜃1 = 𝜋/2 and 𝜃2 = −𝜋/2.
The field then simplifies to become:
𝜇0 𝐼
𝑩=
2𝜋𝐴
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field for a Long, Straight Conductor
Direction
The magnetic field lines are circles concentric with the wire.
The field lines lie in planes perpendicular to the wire.
Magnitude of B is constant on any circle of radius a.
The right-hand rule to determine the direction of B is shown.
Simply grasp the wire with the thumb in the direction of the current. Your fingers
wrap in the direction of the field!
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field for a Curved Wire Segment
Find the field at point O due to the curved wire segment.
I and a are constants.
ds is parallel to r along the line A to A’ and along CC’, so
we only need to work out the field resulting from AC.
We find that
𝑩=
𝜇0 𝐼
𝜃
4𝜋𝑎
, where θ is in radians…
See example 30.2 for the working here…
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field for a Circular Wire Loop
We can use our previous result, namely:
𝜇0 𝐼
𝑩=
𝜃
4𝜋𝑎
to find the field for a circular loop of wire.
For a circular loop of wire, 𝜃 = 2𝜋, and therefore:
𝜇0 𝐼
𝜇0 𝐼
𝑩=
2𝜋 =
4𝜋𝑎
2𝑎
This is the field at the centre of the loop, which has radius a.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Field Lines for a Loop
The left figure, (a), shows the magnetic field lines surrounding a current loop.
The right figure, (b), shows the field lines around a bar magnet.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Force Between Two Parallel Conductors
In the figure to the right, there are two parallel
wires carrying steady currents I1 and I2 .
The field B2 , which is the result of the current in
wire 2, exerts a force on wire 1 of
𝐹1 = 𝐵2 𝐼1 𝑙
𝜇 𝐼
We know that 𝐵2 = 0 2 , from earlier – so we can
2𝜋𝑎
substitute this to get:
𝜇0 𝐼1 𝐼2 𝑙
𝐹1 =
2𝜋𝑎
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Force Between Two Parallel Conductors
𝜇0 𝐼1 𝐼2 𝑙
𝐹1 =
2𝜋𝑎
This equation tells us that:
Parallel conductors that carry current in the same
direction attract one another.
Parallel conductors that carry current in opposing
directions repel one another.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Magnetic Force Between Two Parallel Conductors
This result is often expressed as the magnetic force between two wires, FB.
It can be rearranged to give us the force per unit length:
𝐹𝐵 𝜇0 𝐼1 𝐼2
=
𝑙
2𝜋𝑎
The derivation here assumes that both wires are long compared to their
separation distance, although technically only one wire needs to be long.
The equations accurately describe the forces exerted on each other by a long
wire and a straight, parallel wire of limited length, 𝑙.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Quick Quiz
If, in the figure to the right, we set I1 = 2 A and I2 = 6 A, which is true?
a) F1 = 3 F2
b) F1 = F2 / 3
c) F1 = F2
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Quick Answer
If, in the figure to the right, we set I1 = 2 A and I2 = 6 A, which is true?
c) F1 = F2
Newton’s third law requires that F1 = F2.
Another route to the same answer is to realise that:
𝜇0 𝐼1 𝐼2 𝑙
𝐹1 =
2𝜋𝑎
Gives the same result whether the multiplication is 2x6 or 6x2.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Quick Quiz
A loose spiral spring is hung from the ceiling, and a large current is passed
through it.
The coils move:
a) Closer together
b) Farther apart
c) They do not move at all
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Quick Answer
A loose spiral spring is hung from the ceiling, and a large current is passed
through it.
The coils move:
a) Closer together
The coils act like wires carrying parallel currents in the same direction, and
hence they attract one another.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Definition of the Ampere
Earlier, we introduced the concept of Electric Current, and mentioned the unit is
the ampere (A).
The force between two parallel wires can be used to define the ampere.
When the magnitude of the force per unit length between two long, parallel
N
wires that carry identical currents and are separated by 1m is exactly 2 × 10−7
m
then the current that is carried by each wire is defined to be 1 A.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Sources of the Magnetic Field
Definition of the Coulomb
The SI unit of charge, the coulomb, is actually defined in terms of the ampere.
When a conductor carries a steady current of 1 A, the quantity of charge that
flows through a cross section of the conductor in 1 s is 1 C.
Electricity & Magnetism – Magnetism
21/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Andrew-Marie Ampère
1775 – 1836
French physicist and mathematician
Carried out experiments following Oersted’s
observation that a compass is deflected by an
electric current.
Showed that electricity and magnetism were
linked, and so is often credited as being the
discoverer of Electromagnetism.
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
The Magnetic Field for a Long, Straight Conductor
As we saw earlier, the magnetic field lines are
circles concentric with the wire.
The magnitude of the field is constant on any
circle of radius a.
The right-hand rule for determining the field’s
direction is shown to the right…
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
The Magnetic Field of a Wire
We can use a compass to detect the magnetic
field, reproducing Oersted’s experiment.
When no current flows in the wire, there is no
field due to the current.
The needle will therefore point roughly
towards the Earth’s north geographic pole.
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
The Magnetic Field of a Wire
When the wire carries a current, a compass
needle deflects in a direction tangent to the
circle.
This shows the direction of the magnetic field
produced by the wire.
Just like our old friend,
the iron filings!
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Ampere’s Law
The line integral of B.ds around any closed path equals μ0I, where I is the total
steady current passing through any surface bounded by the closed path.
i.e.
𝑩. 𝑑𝒔 = 𝜇0 𝐼
Ampere’s law describes the creation of magnetic fields by all continuous current
configurations – and is particularly useful when the current configuration has a
high degree of symmetry.
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Ampere’s Law – which way to integrate?
𝑩. 𝑑𝒔 = 𝜇0 𝐼
In order to evaluate the direction you should integrate around a given amperian
loop (i.e. a closed path carrying current):
Put the thumb of your right hand in the direction of the current through the
loop, and your fingers will then curl in the direction you should integrate in.
For more discussion on the background to Ampere’s law, see pages 869-870!
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Quick Quiz
Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed
paths shown in the figure to the right, from
the least to the greatest:
a) a, b, c, d
b) b, d, a, c
c) c, d, b, a
d) c, b, a, d
e) d, c, a, b
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Quick Answer
Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed
paths shown in the figure to the right, from
the least to the greatest:
b) b, d, a, c
Remember:
𝑩. 𝑑𝒔 = 𝜇0 𝐼
Ampere’s law indicates that the value of the
line integral depends only on the net current
through each closed path.
Path b encloses 1 A, path d 3 A, path a 4 A and path c encloses 6 A
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Quick Quiz
Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed
paths shown in the figure to the right, from
the least to the greatest:
a) a, b, c, d
b) b, c, d, a
c) b, d, a, c
d) d, c, a, b
e) The criteria are badly chosen, in this case
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Quick Answer
Rank the magnitudes of 𝑩. 𝑑𝒔 for the closed
paths shown in the figure to the right, from
the least to the greatest:
e) The criteria are badly chosen, in this case
Ranked from least to greatest, first comes b,
but then a = c = d. These three paths all give
the same non-zero value 𝑩. 𝑑𝒔 = 𝜇0 𝐼 since
the size and shape of the paths do not matter.
Path b does not enclose the current, and hence its line integral is zero.
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Field Due to a Long Straight Wire – From Ampere’s Law
Calculate the magnetic field at a distance r from
the centre of a wire that carries a steady current, I.
The current is uniformly distributed through the
cross section of the wire.
Since the wire has a high degree of symmetry, the problem can be categorised as
an Ampere’s Law problem.
For r ≥ R, the result should be the same as that obtained using the Biot-Savart
law.
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Field Due to a Long Straight Wire – From Ampere’s Law
Outside the wire, r > R
𝑩. 𝑑𝒔 = 𝐵. 2𝜋𝑟 = 𝜇0 𝐼
∴ 𝐵=
𝜇0 𝐼
2𝜋𝑟
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Field Due to a Long Straight Wire – From Ampere’s Law
Inside the wire, we need I’, which is the current
Inside the amperian circle:
𝐼′
=
𝜋𝑟 2
𝐼
𝜋𝑅 2
Using this:
𝑩. 𝑑𝒔 = 𝐵 2𝜋𝑟 = 𝜇0 𝐼′
Substitute in for I’, giving us:
𝐵 2𝜋𝑟 =
𝑟2
𝜇0 2 𝐼
𝑅
Electricity & Magnetism – Magnetism
→
𝐵=
𝜇0 𝐼
2𝜋𝑅2
𝑟
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
Field Due to a Long Straight Wire – Results Summary
r<R
𝐵=
r>R
𝐵=
𝜇0 𝐼
2𝜋𝑅 2
𝑟
𝜇0 𝐼
2𝜋𝑟
The field is proportional to r, inside the wire.
The field varies as 1/r outside the wire.
The two equations are equal at r = R.
Much easier to calculate using Ampere’s law rather than the Biot-Savart law!
Electricity & Magnetism – Magnetism
22/08/2013
PHYSICS 1B – Ampere's Law and Magnetic Fields
The Magnetic Field of a Toroid
Find the field at a distance r from the centre
of the toroid, which has N turns of wire.
The wire passes through the loop N times, so
the total current through the loop is N times I.
𝑩. 𝑑𝒔 = 𝐵. 2𝜋𝑟 = 𝜇0 𝑁𝐼
∴
𝜇0 𝑁𝐼
𝐵=
2𝜋𝑟
Electricity & Magnetism – Magnetism
22/08/2013