Chapter 30: Sources of Magnetic fields

Reading assignment: Chapter 30

Homework 30 (due Friday, April 19): QQ1, QQ2, QQ3, QQ4, QQ5, OQ1, OQ2, OQ4, OQ5, OQ6, OQ8, OQ9, OQ10, 1, 2, 3, 5, 20, 21, 23, 39, 45

• Electric currents produce magnetic fields

• Field of a straight current-carrying wire, field of a wire loop, field of a solenoid

• Force between two parallel conductors/wires.

• Tools to calculate/evaluate magnetic fields

• Biot-Savart law

• Ampere’s law

• Gauss’s law for magnetism, magnetic flux

• Magnetic flux

• Magnetism in matter (ferromagnetism, paramagnetism, diamagnetism)

Key concept: Electric currents produce magnetic fields

Observation: Electric currents produce magnetic fields

Right hand rule: Grasp the wire with your right hand so that the thumb points in

the direction of the conventional current (positive); then your fingers will encircle

the wire in the direction of the magnetic field.

• Magnetic field of a long, straight wire is circular around the wire (right hand rule).

• Proportional to I and to 1/r

•It’s strength is given by:

r

IB

2

0

I – current in the wire

r – distance from wire

m0 – permeability of free space

m0 = 4p·10-7 T·m/A

Magnetic field due to an infinitely long, straight wire

Magnetic field lines of a long, straight, current-carrying wire

i-clicker

A battery establishes a steady current around the circuit below. A compass

needle is placed successively at points P, Q, and R to measure the magnetic

field. Rank the strength of the magnetic field from strongest to weakest.

A) P, Q, R. B) Q, R, P. C) R, Q, P. D) P, R, Q. E) Q, P, R

Magnetic field lines produced from a wire loop

• Again, we an use right hand rule to find direction of field

• Looks like field of a tiny magnetic moment (bar magnet)

Magnetic field of a solenoid

• Strong, nearly homogeneous field inside

• Fields from all loops add up

• Resembles field of a bar magnet

0B n I

Field inside solenoid:

I … current in the wire

n = N/L … number of coils per length

m0 … permeability of free space

Magnetic force between two parallel conductors (wires)

Consider two wires carrying currents I1 and I2, separated by a distance a. Each current produces

a magnetic field that is felt by the other.

For example I2 “feels” the field from I1:

0 11 2

IB

a

The force (per unit length) on conductor 2 is:

12BIl

F

0 1 2

2

I IF

l a

Thus, the force on I2 due to the field of I1:

Attractive if currents are in same direction; repulsive if currents are in opposite direction.

White board example

Force between two current carrying wires.

The two wires are 2.0 m long and are 3.0 mm apart and carry a current of 8.0 A.

1) Calculate the force between the wires.

2) Looking at the figure, is the force attractive of repulsive?

aA) AttractiveB) RepulsiveC) No force

The Biot-Savart Law(Used to calculate magnetic fields of currents, e.g., a current-carrying wire)

• Magnetic fields go around the wire – they are perpendicular to the direction of current

• Magnetic fields are perpendicular to the separation between the wire and the point where you measure it - Sounds like a cross product!

02

ˆ

4

I d

r

s rB

034

I d

r

s rB

70 4 10 T m/A

Permeability of free space

Magnetic Field from a Finite Wire

•Magnetic field from a finite straight wire:•Let a be the distance from the wire to point P. •Let x be the horizontal separation

r

I

ds

P

a

x-x1 x2

21

O

0

2

IB

a

01 2sin sin

4

IB

a

for long straight wire

a1 a2

Ampere’s Law•Suppose we have a wire coming out of the plane•Let’s integrate the magnetic field around a closed path

•There’s a new symbol for such an integral•Circle means “over a closed loop”

•The magnetic field is parallel to direction of integration

0

2

IB

a

dB s B ds 0 22

Ir

r

0I

•What if we pick a different path?

dB s cosBds I0

2

Ird

r

0d I B s

ds

cosds rd

ds cosrd

•We have demonstrated this is true no matter what path you take•Wire doesn’t even need to be a straight infinite wire•All that matters is that current passes through the closed Ampere loop

Understanding Ampere’s Law

• If multiple currents flow through, add up all currents that are inside the loop• Use right-hand rule to determine if they count as + or –

• Curl fingers in direction of Ampere loop• If thumb points in direction of current, plus, otherwise minus

• The wire can be bent, the loop can be any shape, even non-planar

0d I B s

6 A

2 A1 A

4 A

7 A

i-clickerThere are currents going in and out of the screen as sketched at right. What is the integral of the magnetic field around the path sketched in purple?A) 0(4 A) B) 0(-4 A) C) 0(12 A)D) 0(-12 A) E) None of the above

• Right hand rule causes thumb to point down• Downward currents count as +, upwards as –

0 4 A 2 A 6 Ad B s

Using Ampere’s Law• Ampere’s Law can be used – rarely – to calculate magnetic fields• Need lots of symmetry – usually cylindrical

A wire of radius a has total current I distributed uniformly across its cross-sectional area. Find the magnetic field everywhere.

I IEnd-on view

• Draw an Ampere loop outside the wire – it contains all the current• Magnetic field is parallel to the direction of this loop, and constant around it• Use Ampere’s Law:• But we used a loop outside the wire, so we only have B for r > a

0I d B s B ds 2 rB0

2

IB

r

Using Ampere’s Law (2)• Now do it inside the wire• Ampere loop inside the wire does not contain all the current• The fraction is proportional to the area

End-on view

2 rB

2

2rI r

I a

a2 2

rI Ir a

0 rI d B s

0

2rI

Br

022

Ir

a

02

0

2

2

I r r aB

Ir a r a

Solenoids• Consider a planar loop of wire (any shape) with a

current I going around it. • Now, stack many such loops

• Treat spacing as very closely spaced• Assume stack is tall compared to size of

loop• Can show using symmetry that magnetic field is

only in vertical direction• Can use Ampere’s Law to show that it is

constant inside or outside the solenoid

00 I d B s 1B L 0 2B L 0

1 2B B

• But magnetic field at infinity must be zero

outside 0B

Field Inside a Solenoid• It remains only to calculate the magnetic field inside• We use Ampere’s law• Recall, no significant B-field outside• Only the inside segment contributes

• There may be many (N) current loops within this Ampere loop

• Let n = N/L be loops per unit length

0 totI d B s

L

inB LtotI NI

0in

NIB

L

in 0B nI

• Works for any shape solenoid, not just cylindrical• For finite length solenoids, there are “end effects”• Real solenoids have each loop connected to the next, like a

helix, so it’s just one long wire

Magnetic Flux

• Magnetic flux is defined exactly the same way for

magnetism as it was for electricityˆ

B dA B n

A cylindrical solenoid of radius 10 cm has length 50 cm and

has 1000 turns of wire going around it. What is the magnetic

field inside it, and the magnetic flux through it, when a

current of 2.00 A is passing through the wire?

Unit of magnetic flux: Tesla·meter2; also called a Weber (Wb)

i-clickerA regular tetrahedron (four sides, all the same) has a cylindrical magnet placed in the middle of the bottom face. There is a total of 0.012 Tm2 of magnetic flux entering the bottom face. What is the total flux from one of the three top faces?A) 0.006 Tm2 B) 0.004 Tm2 C) 0.003 Tm2

D) 0.012 Tm2 E) None of the above

Gauss’s Law for Magnetism• Magnetic field lines always go in circles – there are no magnetic monopole

sources.• For closed surfaces, any flux in must go out somewhere else.

ˆ 0B dA B n

• Flux in bottom must equal total flux out other three sides• Other three sides must have equal flux, by symmetry

i-clicker

A sphere of radius R is placed near a long, straight wire that

carries a steady current I. The magnetic field generated by the

current is B. The total magnetic flux passing through the sphere

is 

)A m0 I)B m0 I/(4 pR2 )

C) 4 pR2/ 0ID) ZeroE) Need more information

A cube of edge length = 2.50 cm is positioned as shown below. A uniform magnetic field

given by B = (5.00 i + 4.00 j + 3.00 k) T exists throughout the region.

(A) Calculate the flux through the shaded region.

(B) What is the total flux through the six faces?

White board example