Today in Physics 217: magnetoquasistatics
Today in Physics 217: magnetoquasistatics Magnetoquasistatics Example magnetoquasistatic calculations with Faraday’s Law Faraday-induced electric fields and the magnetic vector potential When does a uniformly charged spherical shell have nonzero E inside? Flux tubes on the Sun, seen by NASA’s TRACE satellite. 22 November 2002
Physics 217, Fall 2002
1
Magnetoquasistatics With Faraday’s Law, we can pose lots of new problems in which one generates time-variable magnetic fields, and calculates electric fields. We have derived formulas for B generated by lots of current distributions and we can use… But can we? All those calculations assume steady currents, and give steady magnetic fields. Thus, to use them straight away with time-varying currents gives magnetic fields with the same time dependence everywhere in space. This can’t be right! If I look far enough away from the currents I should see a magnetic field that depends on what the current was a little while ago, because the electromagnetic information of what the current is, can propagate to that point no faster than the speed of light. 22 November 2002
Physics 217, Fall 2002
2
Magnetoquasistatics (continued) So any calculation we do in this manner gives only an approximate solution, good only if the current doesn’t fluctuates very fast or if we’re not calculating the field very far away. If τ is the time over which the current changes and r is the distance away from the currents at which we want to calculate the field, then we need to stay within Magnetoquasistatic r cτ . approximation
You will learn how to do it right next semester in PHY 218, using retarded fields and retarded potentials. But for now we’ll assume we’re to be working within the quasistatic approximation. 22 November 2002
Physics 217, Fall 2002
3
Magnetoquasistatics (continued) For instance, consider Example 7.9: “An infinitely long straight wire carries a current I(t). Determine the induced electric field, a distance s from the wire.” The result is 2 dI s E= ln zˆ → ∞ . s →∞ c dt s0 The fact that it blows up as s → ∞ is an indication of the violation of the magnetoquasistatic approximation at distances too great for the field to be in sync with the current. See Example 10.2 for the sorts of things that are necessary to get the right answer. (See it next semester, though.) 22 November 2002
Physics 217, Fall 2002
4
Electric field from a variable solenoidal current An infinite solenoid with radius a and n turns per unit length carries a time-dependent current I(t) in the φˆ direction. Find the electric field at a distance s from the axis, inside and out. The magnetic field is zero outside and 4π nIzˆ B= c inside, so the flux through a circular loop with radius s z s is 4π 2 nI s π (s < a) a c ΦB = 4π nIπ a2 ( s > a ) c 22 November 2002
Physics 217, Fall 2002
5
Electric field from a variable solenoidal current (continued) And
so
1 dΦ B v∫ E ⋅ dA = − c dt
4π 2 s 2 n dI − dt c2 ⇒ E2π s = 2 2 4π a n dI − dt c2
(s < a) (s > a)
2π sn dI ˆ − 2 dt φ ( s < a ) c E= 2 π a n dI 2 − ˆ (s > a) φ c 2 s dt
For MKS units, change one of the factors of 1/c to µ0 4π , and the other to unity. 22 November 2002
Physics 217, Fall 2002
6
Breaking a wire A square conducting loop, with side a and resistance R, lies a distance a from an infinite straight wire that carries current I. Suppose that we break the wire, so that I drops abruptly to zero. What total charge passes a given point in the loop during the time this current a flows, and in what direction does the induced current in a the square loop flow? a
z
Field from the straight infinite current: 2I ˆ . B= φ cs 22 November 2002
Physics 217, Fall 2002
7
Breaking a wire (continued) So the flux of B through the square loop is 2 Ia ΦB = c
2a
∫ a
2 Ia ds 2 Ia 2a ln s a = ln 2 = s c c
, and
1 dΦ B 2 a ln 2 dI dQ R=− E = I loop R = =− dt c dt c 2 dt
.
Note that Q depends on the current in the loop, not I. And it’s Q we want: Q
∫
Q = dQ′ = − 0
2 a ln 2 2
c R
0
∫ dI ′ = I
2 Ia ln 2 c2 R
µ0 Ia ln 2 = in MKS . 2π R
It doesn’t matter how fast the current shuts off… 22 November 2002
Physics 217, Fall 2002
8
Breaking a wire (continued) As for the direction of the induced current: Before the wire is cut, the field points out of the page. Shutting off the current decreases this field. The current in the loop will flow so as to oppose this change (Lenz’s Law.) If the induced current flows counterclockwise, it will produce a magnetic field in the same a direction as the (decreasing) field from the wire, which is a what it takes to oppose the a z change.
22 November 2002
Physics 217, Fall 2002
9
Faraday-induced electric fields and the magnetic vector potential In magnetostatics,
4π 1 J ( r ′ ) × rˆ ′. J , B= ∫ d τ —⋅B = 0 , —×B = c c r2 For induced electric fields and ρ = 0, we can substitute 1 ∂B , B⇔E , J⇔− 4π ∂t and get 1 ∂B 1 ∂ B ( r ′, t ) × rˆ ′. , E=− d τ —⋅E = 0 , —× E = − ∫ 4π c ∂t c ∂t r2
22 November 2002
Physics 217, Fall 2002
10
Faraday-induced electric fields and the magnetic vector potential (continued) Furthermore,
—⋅ A = 0 , —× A = B ,
4π J. so A depends on B in the same way that B depends on c ˆ ′ B r , × t r ( ) 1 Thus A= dτ ′ , and 2 4π r 1 ∂A E=− . c ∂t
∫
1 ∂ 1 ∂B Check: — × E = − —× A = − c ∂t c ∂t
(Faraday's Law again).
Leave off the factor of 1/c for MKS.
22 November 2002
Physics 217, Fall 2002
11
When does a uniformly charged spherical shell have nonzero E inside? A spherical shell with radius R carries a uniform charge density σ. It spins about the z axis at angular velocity ω ( t ) that changes with time. Find the electric field inside and outside the sphere. The week before last (Example 5.11, lecture, 11 November 2002), we worked out the magnetic vector potential from a spinning, uniformly charged spherical shell, and got 4π Rσω ˆ 3c r sin θ φ ( r < R ) , A(r ) = 4 4 R π σω sin θ ˆ (r > R). φ 3 3c r A is time dependent if ω is. 22 November 2002
Physics 217, Fall 2002
12
When does a uniformly charged spherical shell have nonzero E inside? (continued) So there are two contributions to the electric field: electrostatic, and induced. The static part is, of course,
0 ( r < R ) , Estatic ( r ) = 4π R 2σ ˆ (r > R). r r2 Using the E-A relation we just derived, the induced part is 4π Rσ dω ˆ (r < R) , r sin θ φ − 2 dt 1 ∂A 3c Eind ( r , t ) = − = c ∂t 4π R 4σ dω sin θ ˆ (r > R). φ − 3c 2 dt r 3 22 November 2002
Physics 217, Fall 2002
13
When does a uniformly charged spherical shell have nonzero E inside? (continued) So the total electric field is 4π Rσ dω ˆ (r < R) φ − θ sin r 2 dt 3c E (r ,t) = , or 2 4 4π R σ rˆ − 4π R σ dω sin θ φ ˆ (r > R) r 2 3c 2 dt r 3 µ0 Rσ dω ˆ − 3 dt r sin θ φ ( r < R ) = 2 in MKS. 4 R σ rˆ − µ0 R σ dω sin θ φ ˆ ( r > R ) ε 0 r 2 dt r 3 3 It’s nonzero inside, if the angular velocity changes with time. 22 November 2002
Physics 217, Fall 2002
14
Physics 217, Fall 2002
1
Magnetoquasistatics With Faraday’s Law, we can pose lots of new problems in which one generates time-variable magnetic fields, and calculates electric fields. We have derived formulas for B generated by lots of current distributions and we can use… But can we? All those calculations assume steady currents, and give steady magnetic fields. Thus, to use them straight away with time-varying currents gives magnetic fields with the same time dependence everywhere in space. This can’t be right! If I look far enough away from the currents I should see a magnetic field that depends on what the current was a little while ago, because the electromagnetic information of what the current is, can propagate to that point no faster than the speed of light. 22 November 2002
Physics 217, Fall 2002
2
Magnetoquasistatics (continued) So any calculation we do in this manner gives only an approximate solution, good only if the current doesn’t fluctuates very fast or if we’re not calculating the field very far away. If τ is the time over which the current changes and r is the distance away from the currents at which we want to calculate the field, then we need to stay within Magnetoquasistatic r cτ . approximation
You will learn how to do it right next semester in PHY 218, using retarded fields and retarded potentials. But for now we’ll assume we’re to be working within the quasistatic approximation. 22 November 2002
Physics 217, Fall 2002
3
Magnetoquasistatics (continued) For instance, consider Example 7.9: “An infinitely long straight wire carries a current I(t). Determine the induced electric field, a distance s from the wire.” The result is 2 dI s E= ln zˆ → ∞ . s →∞ c dt s0 The fact that it blows up as s → ∞ is an indication of the violation of the magnetoquasistatic approximation at distances too great for the field to be in sync with the current. See Example 10.2 for the sorts of things that are necessary to get the right answer. (See it next semester, though.) 22 November 2002
Physics 217, Fall 2002
4
Electric field from a variable solenoidal current An infinite solenoid with radius a and n turns per unit length carries a time-dependent current I(t) in the φˆ direction. Find the electric field at a distance s from the axis, inside and out. The magnetic field is zero outside and 4π nIzˆ B= c inside, so the flux through a circular loop with radius s z s is 4π 2 nI s π (s < a) a c ΦB = 4π nIπ a2 ( s > a ) c 22 November 2002
Physics 217, Fall 2002
5
Electric field from a variable solenoidal current (continued) And
so
1 dΦ B v∫ E ⋅ dA = − c dt
4π 2 s 2 n dI − dt c2 ⇒ E2π s = 2 2 4π a n dI − dt c2
(s < a) (s > a)
2π sn dI ˆ − 2 dt φ ( s < a ) c E= 2 π a n dI 2 − ˆ (s > a) φ c 2 s dt
For MKS units, change one of the factors of 1/c to µ0 4π , and the other to unity. 22 November 2002
Physics 217, Fall 2002
6
Breaking a wire A square conducting loop, with side a and resistance R, lies a distance a from an infinite straight wire that carries current I. Suppose that we break the wire, so that I drops abruptly to zero. What total charge passes a given point in the loop during the time this current a flows, and in what direction does the induced current in a the square loop flow? a
z
Field from the straight infinite current: 2I ˆ . B= φ cs 22 November 2002
Physics 217, Fall 2002
7
Breaking a wire (continued) So the flux of B through the square loop is 2 Ia ΦB = c
2a
∫ a
2 Ia ds 2 Ia 2a ln s a = ln 2 = s c c
, and
1 dΦ B 2 a ln 2 dI dQ R=− E = I loop R = =− dt c dt c 2 dt
.
Note that Q depends on the current in the loop, not I. And it’s Q we want: Q
∫
Q = dQ′ = − 0
2 a ln 2 2
c R
0
∫ dI ′ = I
2 Ia ln 2 c2 R
µ0 Ia ln 2 = in MKS . 2π R
It doesn’t matter how fast the current shuts off… 22 November 2002
Physics 217, Fall 2002
8
Breaking a wire (continued) As for the direction of the induced current: Before the wire is cut, the field points out of the page. Shutting off the current decreases this field. The current in the loop will flow so as to oppose this change (Lenz’s Law.) If the induced current flows counterclockwise, it will produce a magnetic field in the same a direction as the (decreasing) field from the wire, which is a what it takes to oppose the a z change.
22 November 2002
Physics 217, Fall 2002
9
Faraday-induced electric fields and the magnetic vector potential In magnetostatics,
4π 1 J ( r ′ ) × rˆ ′. J , B= ∫ d τ —⋅B = 0 , —×B = c c r2 For induced electric fields and ρ = 0, we can substitute 1 ∂B , B⇔E , J⇔− 4π ∂t and get 1 ∂B 1 ∂ B ( r ′, t ) × rˆ ′. , E=− d τ —⋅E = 0 , —× E = − ∫ 4π c ∂t c ∂t r2
22 November 2002
Physics 217, Fall 2002
10
Faraday-induced electric fields and the magnetic vector potential (continued) Furthermore,
—⋅ A = 0 , —× A = B ,
4π J. so A depends on B in the same way that B depends on c ˆ ′ B r , × t r ( ) 1 Thus A= dτ ′ , and 2 4π r 1 ∂A E=− . c ∂t
∫
1 ∂ 1 ∂B Check: — × E = − —× A = − c ∂t c ∂t
(Faraday's Law again).
Leave off the factor of 1/c for MKS.
22 November 2002
Physics 217, Fall 2002
11
When does a uniformly charged spherical shell have nonzero E inside? A spherical shell with radius R carries a uniform charge density σ. It spins about the z axis at angular velocity ω ( t ) that changes with time. Find the electric field inside and outside the sphere. The week before last (Example 5.11, lecture, 11 November 2002), we worked out the magnetic vector potential from a spinning, uniformly charged spherical shell, and got 4π Rσω ˆ 3c r sin θ φ ( r < R ) , A(r ) = 4 4 R π σω sin θ ˆ (r > R). φ 3 3c r A is time dependent if ω is. 22 November 2002
Physics 217, Fall 2002
12
When does a uniformly charged spherical shell have nonzero E inside? (continued) So there are two contributions to the electric field: electrostatic, and induced. The static part is, of course,
0 ( r < R ) , Estatic ( r ) = 4π R 2σ ˆ (r > R). r r2 Using the E-A relation we just derived, the induced part is 4π Rσ dω ˆ (r < R) , r sin θ φ − 2 dt 1 ∂A 3c Eind ( r , t ) = − = c ∂t 4π R 4σ dω sin θ ˆ (r > R). φ − 3c 2 dt r 3 22 November 2002
Physics 217, Fall 2002
13
When does a uniformly charged spherical shell have nonzero E inside? (continued) So the total electric field is 4π Rσ dω ˆ (r < R) φ − θ sin r 2 dt 3c E (r ,t) = , or 2 4 4π R σ rˆ − 4π R σ dω sin θ φ ˆ (r > R) r 2 3c 2 dt r 3 µ0 Rσ dω ˆ − 3 dt r sin θ φ ( r < R ) = 2 in MKS. 4 R σ rˆ − µ0 R σ dω sin θ φ ˆ ( r > R ) ε 0 r 2 dt r 3 3 It’s nonzero inside, if the angular velocity changes with time. 22 November 2002
Physics 217, Fall 2002
14
