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Year 2 Electromagnetism > Ampere-Maxwell Law > Flashcards

Ampere-Maxwell Law Flashcards

1
Q

Biot-Savart Law for a Point Charge

A

|B(|r) = μ0/4π * (q|v x ^r)/r²
-a charge q moving at speed v produces a magnetic field of magnitude B

|F(|r) = q * [ |E(|r) + (|v x |B(|r)) ]
-a charge q experiences a force F when moving through an electric field of magnitude F and at velocity v through a magnetic field of magnitude B

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2
Q

Derivation of the Biot-Savart Law for a Current Carrying Element

A

-starting with the Biot-Savart law for a point charge
|B(|r) = μ0/4π * (q|v x ^r)/r²
-a small element of the wire contains charge dq moving at velocity v
d|B(|r) = μ0/4π * (dq|v x ^r)/r²
-the veocity of the charge is given by d|l/dt where |l is a small element of the wire
-current is given by I=dq/dt
dq|v = dq* d|l/dt = dq/dt * d|l = I d|l
-sub in to obtain the Biot-Savart Law
d|B(|r) = μ0/4π * (Id|l x ^r)/r²

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3
Q

Which direction is d|l taken to be in?

A

-the direction of positive current flow

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4
Q

How do you calculate the magnetic field produced due to a current loop, solenoid or straight wire?

A
  • start with the Biot-Savart law for a current carrying element that produces a magnetic field
  • integrate over all current carrying elements in the wire
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5
Q

Magnetic Field of an Infinitely Long Straight Wire

Equation

A

B = μoI / 2πR

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6
Q

Magnitude of the Magnetic Field of a Straight Wire

Derivation

A

-to find the magnetic field at a point P which is y above the wire, start with the Biot-Savart law for a charge element
d|B(|r) = μo/4π * (Id|l x ^r)/r²
-where r is the vector between d|l and P, calculate the magnitude
dB = |d|B(|r)| = μ0/4π * (I | d|l x ^r |)/r²
-choose the x axis in the direction of current flow
| d|l x ^r | = | dx^ix^r| = dx| ^i x ^r | = dx sinφ
-where φ is the angle between r and the positive x axis
-sub in
dB = μo/4π * (I dx sinφ) / r²
-sub in for θ, where θ is the angle between r and y
dB = μo/4π * (I dx cosθ) / r²
->x = ytanθ
-differentiate
dx = y sec²θ dθ
secθ = r/y
-sub in
dB = μo/4π * I/y * cosθ dθ
-integerate between θ1 and -θ2
B = μo/4π * I/y * [sinθ1 + sinθ2]

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7
Q

Magnitude of Magnetic Field due to a Straight Wire

Equation

A

B = μo/4π * I/y * [sinθ1 + sinθ2]

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8
Q

Force on a Current Carrying Wire in a Magnetic Field

Derivation

A

-let the drift velocity of charges in the wire be vd, the cross sectional area of the wire is A, each charge carrier has charge q, and the charge density per unit volume is n
-consider a wire of length l carrying a current I in a magnetic field B
-the number of charge carriers in length l is:
nAl
-each individual charge carrier, q , experiences a force:
|F = q|vd x |B
-total force on the wire is the force on one charge carrier multiplied by the number of charge carriers
|F = (nAl)q |vd x |B = qnAvd(|l x |B)
-recall I = qnAvd
|F = I |l x B

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9
Q

Force on a Current Carrying Wire

Equation

A

|F = I |l x B

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10
Q

Definition of the Ampere

A

-the ampere is the current that if it flows in two straight parallel infinitely long narrow wires placed one meter apart in a vacuum will produce a force of 2x10^(-7) newtons per meter length of wire on each wire

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11
Q

Force Experience by a Wire Due to Another Current Carrying Wire
Equation

A

dF2/dl2 = 2 * μo/4π * I1I2/R

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12
Q

Force Experienced by a Wire Due to Another Current Carrying Wire
Derivation

A

-force experienced by wire 2 due to the magnetic field of wire 1 is:
dF2 = B1I2dl2
-the magnetic field of wire 1 is given by:
B1 = μoI1/2πR
-sub in
dF2 = μoI1/2πRI2dl2
-rearrange for force per unit length on wire 2
dF2/dl2 = 2 * μo/4π * I1I2/R

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13
Q

Ampere’s Law

Integral Form

A

∮|B . d|l = μo Ic

-an electric current, Ic, through a surface produces a circulating magnetic field around the path that bounds that surface

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14
Q

When can Ampere’s Law be used?

A
  • Ampere’s Law is only valid if the following two conditions are met:
    i) the current has to be constant, i.e not time dependent
    ii) the current can not be spatially dependent, i.e. no start or end point so you need a LONG wire
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15
Q

Ampere’s Law

Differential Form - Derivation

A

-starting with the definition of curl (where limit is taken as ΔA->0
(|∇x|B) . ^n = lim 1/ΔA * ∮|B . d|L
-using the differential form ∮|B . d|L = μo Ic :
(|∇x|B) . ^n = lim 1/ΔA * μo Ic
-take out constants:
(|∇x|B) . ^n = μo * lim Ic/ΔA
-sub in for current density, Jn
(|∇x|B) . ^n = μo Jn
-normal component of (|∇x|B) is always equal to the normal component of Jn so we can say that:
|∇x|B = μo |Jn

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16
Q

Ampere’s Law

Differential Form - Equation

A

|∇x|B = μo |Jn

-a circulating magnetic field is produced by an electric current

17
Q

Current Density

A

Jn = lim Ic/ΔA
-where the limit is taken as ΔA -> 0

|J = Jx ^i + Jy ^j + Jz ^k

-current density is the current per unit are through a surface as the area tends to zero, i.e. the current through a point

18
Q

Current

Current Density Definition

A

-current can be thought of as the flux of the current density through a surface
-or the amount of charge passing through unit area perpendicular to the flow per second
-only current parallel to the normal of the surface contributes too the flux through the surface
Ic = ∫|J . ^n dA

19
Q

Description of Problems With Ampere’s Law

A
  • for a finite wire the surface considered must be exactly at the centre for symmetry arguments to hold
  • Ampere’s Law cannot distinguish between the magnetic field of an infinite wire and a finite wire, it does not agree with the Biot-Savart Law for a finite wire
  • it also gives two different results for two different surfaces bounded by the same contour for a capacitor since a capacitor demonstrates discontinuous current flow
20
Q

Ampere-Maxwell Law - Derivation

Charge as a Function of Charge Density

A

-consider a fixed volume of space V with a charge density ρ
-the total charge Q in that volume is given by:
Q = ∫ρdV

21
Q

Divergence of Current Density

Equation

A
  • ∂ρ/∂t = |∇ . |J
22
Q

Divergence of Current Density

Description

A

-divergence of the current density at a point in space is equal to the rate of reduction of charge density with respect to time at that point

23
Q

What is displacement current density?

A

-from the altered version of Ampere’s law, we have:
|∇ . |Jd = ∂ρ/∂t
-now take Gauss’s Law for electric fields:
|∇ . |E = ρ / εo
-differentiate both sides with respect to t
∂/∂t (|∇ . |E) = (1/εo) * ∂ρ/∂t
-rearrange
|∇ . εo ∂|E/∂t = ∂ρ/∂t
-recalling |∇ . |Jd = ∂ρ/∂t, gives:
|Jd = εo * ∂|E/∂t

24
Q

Displacement Current

Derivation

A 
-starting with Q=CV and C = εo A/d
Q = εo A/d V = εo V/d A
-since V/d = E
Q = εo E A
-differentiate with respect to time
∂Q/∂t = εo ∂E/∂t A
-by definition, I = ∂Q/∂t
I = εo ∂E/∂t A
-the definition of displacement current density 
|Jd=εo ∂|E/∂t, gives:
I = Jd A = Id
-this derivation assumes that the area and orientation of the plates is constant
25
Q

Displacement Current

Description

A

-in reality there is nothing flowing between the two plates of the capacitor
-the ‘imaginary’ displacement current is ‘flowing’ and is equal to the actual current in the wires
-it is equal to:
εo ∂/∂t ∫|E . ^n dA

26
Q

Ampere-Maxwell Law - Differential Form

Equation

A

|∇ x |B = μo (|J + εo ∂|E/∂t)

27
Q

Ampere-Maxwell Law - Differential Form

Description

A

-a circulating magnetic field is produced by an electric field that changes with time

28
Q

Ampere-Maxwell Law - Integral Form

Equation

A

∫|B.d|l = μo (Ic + εo ∂/∂t ∫|E . ^n dA)

  • R.H.S.- circulation of the magnetic field around a path C
  • L.H.S. - covers the two sources of the magnetic field, a steady conduction current, Ic, and a changing electric flux through any surface S bounded by C
29
Q

Ampere-Maxwell Law - Integral Form

Description

A
  • an electric current or a changing electric flux through a surface produces a circulating magnetic field around any path that bounds that surface
  • OR a magnetic field is produced along a path if any current is enclosed by the path or if the electric flux through any surface bounded by the path changes over time
30
Q

Ampere-Maxwell Law - Integral Form

Solvable Problems

A

1) Given information on enclosed current or change in electric flux, determine the circulation of the magnetic field
2) In highly symmetrical situations you can also determine the magnitude of the magnetic field using a ‘Special Amperian Loop’ along which |B| is constant

31
Q

Curl of a Magnetic Field

A
  • magnetic field lines circle back on themselves so curl must be non-zero at some points in a magnetic field
  • in magnetic fields the locations of non-zero curl are exactly the locations at which current is flowing or an electric field is changing
  • it is zero elsewhere even where the field appears ‘curved’
32
Q

Relationship Between |J, I and S

A

-if |J is uniform and perpendicular to S:
I = |J| * surface area

-if |J is uniform and at an angle to S:
I = (|J . ^n) * surface area

-if |J is nonuniform and at variable angle to S:
I = ∫ |J . ^n dA

33
Q

Displacement Current Density

A

εo ∂|E/∂t has units of A/m²

  • doesn’t represent actual current as there is no physical movement of charge
  • a changing electric field produces a changing magnetic field even when no charges are present and no physical current flows
34
Q

Ampere-Maxwell Law - Differential Form

Solvable Problems

A

1) Given an expression for the vector magnetic field, determine the electric current density
2) Given an expression for the vector magnetic field determine the displacement current

Year 2 Electromagnetism (9 decks)

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