Chapter 12 Magnetic Fields Due to Currents

In this chapter we will explore the relationship between an electric current and the magnetic field it generates in the space around it. We will follow a two-prong approach, depending on the symmetry of the problem.

For problems with low symmetry we will use the law of Biot-Savart in combination with the principle of superposition. (brief)

For problems with high symmetry we will introduce Ampere’s law.

Both approaches will be used to explore the magnetic field generated by currents in a variety of geometries (straight wire, wire loop, solenoid coil, toroid coil)

We will also determine the force between two parallel current carrying conductors. We will then use this force to define the SI unit for electric current (the Ampere)

We will have a short discussion of Magnetization in Materials

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r

Source of Magnetic FieldsScientists in the 19th century(Orsted, Ampere, Biot-Savart) realized through simple experiments that a current carrying wire produces a magnetic field.The shape of the field around a wire could be measured by a set of compasses. The field produced

by a straight wire are concentric circles about the wire, falling off in strength as 1r

,

where r the distance from the wire.

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Bio - Savart LawBiot-Savart proposed a mathematical descrition of the magnetic field that canbe used for wires of any shape. It is called the Biot-Savart Law. Each currentsegment id

!ℓ along the total length of the wire is integrated to produce the field

at a point !r . #r is the unit vector along !r .

d!B =

µ0

4π i d!ℓ× r̂r 2

This integral can bedone in simple cases,Professionalsgenerally use numerical approaches.

Notice that !B, i!ℓ, and r̂

obey a right-hand rule.

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Consider the wire element of length ds shown in the figure.The element generates at point P a magnetic field of

magnitude dB =µoi4π

dssinθr 2 Vector d

!B is pointing

into the page. The magnetic field generated by the whole wire is found by integration.

B = dB = 2 dB0

∞

∫ =−∞

∞

∫µoi2π

dssinθr 2

0

∞

∫r = s2 + R2 sinθ = sinφ = R / r = R / s2 + R2

B =

µoi2πR

Biot Savart Law for the straight Wire

d!B =

µ0

4π i d!ℓ× r̂r 2

ϕ

B =µoi2π

Rds

s2 + R2( )3/2 =0

∞

∫µoi

2πRs

s2 + R2

⎡

⎣⎢

⎤

⎦⎥

0

∞

=µoi

2πR

dx

x2 + a2( )3/2 = x

a2 x2 + a2∫ sin(θ ) = sin(π −θ )4

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6

https://www.khanacademy.org/science/in-in-class10th-physics/in-in-magnetic-effects-of-electric- current/magnetic-field-due-to-current-carrying-loops-and-solenoids/v/magnetic-field-due-to- current-carrying-loop

https://www.khanacademy.org/science/in-in-class10th-physics/in-in-magnetic-effects-of-electric- current/magnetic-field-due-to-current-carrying-loops-and-solenoids/v/magnetic-fields- through-solenoids

Khan Academy clips on magnetic loops and solenoids.

Ampere's Law (1826)Ampere found a form of the the Biot-Savart Law, which in situations that have highsymmetry, is simpler to apply. Gauss's Law states

that the diverging flux E ⋅dA∫∫ of a vector E

field thru an enclosing surface is equal to the

charges enclosed qenc / ε . E ⋅dA!∫∫ = qenc / ε0 !B ⋅d!s"∫ = µoienc

Ampere's Law states the closed path integral !B ⋅d!s"∫ of

!B along closed path

d!s is equal to the net current enclosed in the closed path. !B ⋅d!s"∫ = µ0ienc

SignMattersIf the integration d!s is CCW then currents ⊙ = positive ⊗=negativeIf the integration d!s is CW then currents ⊙ = negative ⊗=positiveThat is, positive current ⊙ always po int s in the direction of the right hand ruleas defined by the direction of d!s. 7

Try this:

The net current enclosed in these Amperean Loops are:

10A !

30A ! 20A ✖

10A !

A

B

C A: (-10-30+20-10) A CW positive in B: (-10-30) A CW positive in C: (+30-20) A CCW positive out

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Applying Ampere's Law : 1) All currents must be steady and do not change with time. 2) Only currents inside the closed path are taken in to account

and contribute to the magnetic field.3) The sign of the current follows the direction of the right hand rule.4) A path is picked where |

!B| is constant or 0 and

!B ⋅d!s is simple to evaluate.

B ds

+i

B ds

+i +i +i +i

wire solenoid

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Ampere’s Law for a Straight Wire (again)

B is a constant along the path of integration and can be taken out of the integral ⎰Bds=B⎰ds=B2πR !

B =

µoi2πR

I

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Ampere’s Law for a Solenoid of N turns

Ampere’s Loop

•  We choose the path so b , c, and d is far from the coil where the field is weak or near zero (Bb,c,d,=0). •  On the path a B and ds are parallel so B"ds = Bds •  B =~constant inside the coil ! •  The integral along side a is⎰B"ds = ⎰Bds= B L •  The enclosed current Ienc = +N i •  From Ampere’s Law B L = µ0N i ->

Ampere’s Law ⎰B"ds =µ0Ni

Bi=µ0Ni / L

L

i a

b

c

c

! ! ! ! ! ! ! ! ! !

✖ ✖ ✖ ✖ ✖ ✖ ✖ ✖

Β

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Ampere’s Law for a Toroid

Ampere’s Loop

•  B is constant along Ampere’s Loop. •  Also B || ds so =B ds cosθ = B ds •  The integral along Ampere’s Loop ⎰B"ds = B 2πr •  Ienc = N i •  From Amperes Law B 2πr = µ0N i

Ampere’s Law ⎰B"ds =µ0Ni

B=µ0Ni / 2πr

B=µ0Ni / 2πr

12

Summary

Bwire =

µ0I2πr

Bsolenoid =µ0NI

L= µ0n I

B =µ0NI2πr

N turns

n = N / ℓturns / meter

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B2 =

µ0I2

2ϖa

F12

F12

F12

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Magnetization inMaterials :!

M( A / m)• Materials display the properties of Paramagnetism, Diamagetism, and Ferromagnetism.•These properties are due to permanent and induced magnetic dipoles

!µ in the material.

iThe magnetization of a sample is expressed as !M=1/V

!µi∑ , the vector sum of dipoles

per unit volume. This can be re-written as !M=n

!µavg where n=

Ndipoles

V=ρNA

mmole

and

µavg = averagedipolemoment. M has units of A/m

i The magnetic field due to magnetization is !BM =µ0

!M .

i The average dipole moment in a material is of

interest to scientists: !µavg =

!Mn

!µavg =

!BM

µ0n

i!BMneeds to be measured in the sample by placing it in an external magnetic field

!B0.

i The induced !BM will be proportional to the applied field

!B0 ,

!BM = χ

!B0 .

χ = magnetic susceptibility of thematerial

Magnetic Domains in a Material

15 M Bo B=Bo+BM

Magnetic SusceptibilityχiThemagnetic susceptibiliy χ measures how much a material will become magnetized in

anapplied magnetic field !Bo.

!BM = χ

!Bo

i Whenan external field !Bois applied to a sample the net field produced is the sum of

external and inside fields. !B =!Bo +

!BM = (1+ χ )

κ"#$

!Bo κ = 1+ χ = permeabilltyconstant.

!B=κ

!Bo

iWe canmeasure the magnetic susceptibility χ by measuring κ =

!B !

Bo

.

B0 B0++ B0=0

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Hysteresis Curve of a Ferromagnet

• In Figure 12.26 we see the magnetic hysteresis curve for a ferromagnet. •  The ferromagnet will saturate at about 1.5 T. Further increase of Bo will not increase B. • Once the magnetic field Bo is applied the ferromagnet retains a permanent hysteresis pattern. • Even at Bo=0 there is permanent magnetization B. • The permanent magnetization may be destroyed by heating or even mechanical shock.

1.5T

(10-4 T)

B0 = µ0nI ~ 10−4T

B =κ B0 ~ 1.5T17

saturation B field

No hysterisis

Little or No hysterisis

http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/magprop.html

Strong hysterisis

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χ<0

χ>0

χ>>0

filled shells

partially filled shells

magnetic domains Co, Ni, Gd

Try this:

(a)κ M !BB0

= 1.4T6.5x10−4T

= 2300 χ ~κ M

(b) M = (1/ µ0 )B = 1.4T4π x10−7T ⋅m / A

= 1.1x106 A / m

M = µi∑ / V = nµFe

Let n=N/V the number of iron atoms contributing

n =ρFeN A

m= (7.85x103kg / m3 )(6.02x1023 atoms / mole)

0.0559kg / mole

n = 8.45x1028 atoms / m3

µFe = M / n µFe = 1.3x10−23 J / T atomic physics

IRON CORE SOLENOID

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Nℓ

STOP HERE

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Magnetic Field in a Gap Magnetic Shielding

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Magnetic Flux Return (Circuit)

Estimate the magnetic field B in the gap assuming all field lines produced in the coils are captured by the iron (κ=1200). and pass through the gap. All dimensions in centimeters.

Bgap =κµ0( Nℓ

)I +κµ0( Nℓ

)I " 2κµ0nI = 2×1200× 4π x10−7 × (500 / 0.5)×10A = 30.2T

Much to high. Something wrong!We need to usethe formalism of Magnetic Circuits!

Try This:

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