Today in Physics 217: inductance
Today in Physics 217: inductance Mutual inductance How mutual inductance can help with magnetic flux calculations: examples of coupled solenoids and coupled circular loops. Self inductance and back EMF Units Calculation of inductance
25 November 2002
Physics 217, Fall 2002
dA 2
B1
r
I
dA 1
1
Mutual inductance Consider two coupled wire loops, arranged such that a current through one of them produces a magnetic field that has flux through the other one. For instance,
Φ B2 = ∫ B1 ⋅ da2 = ∫ ( — × A1 ) ⋅ da2 =v ∫ A1 ⋅ dA 2 But
A1 =
Φ B2
I1 c
I1 = c
dA 2
.
dA 1 , so r dA 2 ⋅ dA 2 ∝ I1 r
r
v∫
v∫
B1
;
I
dA 1
the proportionality factor is a constant that depends only upon the geometry of the loops. 25 November 2002
Physics 217, Fall 2002
2
Mutual inductance (continued) Now suppose we exchange the roles of the loops, and run a current through loop #2. We’d get the same result with just the 1s and 2s switched: I 2 dA 2 ⋅ dA 2 Φ B1 = ∝ I2 . r c
v∫
The proportionality factor is the same in the two cases. Define it as follows: Φ B1 = cMI 2 , Φ B2 = cMI 1 , where M=
1 c2
v∫
dA 2 ⋅ dA 2 r
µ0 = 4π
v∫
dA 2 ⋅ dA 2 in MKS r
is called the mutual inductance of the loops. 25 November 2002
Physics 217, Fall 2002
3
How mutual inductance can help with magnetic flux calculations This simple exercise has a profound conclusion: no matter what the shape of two loops, the flux through loop 2, resulting from current I in loop 1, is exactly the same as the flux through loop 1 due to current I in loop 2. This occasionally is useful in calculations of flux and related quantities: switching the currents and fluxes for the calculation, and then switching back, can save a lot of trouble. Consider Example 7.10: a short solenoid (length A, radius a, n1 turns per unit length) and a very long one (b, n2) are coaxial. Current I flows in the short one. What is the flux through the long one? 25 November 2002
Physics 217, Fall 2002
4
Coupled solenoids This would be difficult if calculated the way it’s stated, but easy if one instead runs the same current through the outer coil and calculates the flux through the inner: ΦB =
4π n2 I × n1 Aπ a2 c
.
a b
25 November 2002
Physics 217, Fall 2002
5
Coupled circular loops A small loop of wire (radius a) lies a distance z above the center of a large loop (radius b). The planes of the two loops are parallel, and perpendicular to the common axis. R= (a) Suppose current I flows in the big z2 + b 2 loop. Find the flux in the little loop. (b) Suppose current I flows in the little loop. Find the flux through the big loop. (c) What is the mutual inductance?
25 November 2002
Physics 217, Fall 2002
a
z
b
6
Coupled circular loops (continued) (a) We saw long time ago (Example 5.6, lecture notes for 4 November 2002) that Ib b2 2π B= zˆ , 3 2 c z2 + b 2 R=
(
)
∫
25 November 2002
(
z
z2 + b 2
so if the loop is small enough to consider B to be constant over its extent,
2π 2 Φ Ba = Bb ⋅ daa ≅ Bb Aa = c
a
b
b 2 a2 z2 + b
Ib 3 2 2
Physics 217, Fall 2002
)
.
7
Coupled circular loops (continued) (b) Supposing that the small loop is small enough to treat as a dipole, 2 m cos θ m sin θ ˆ ˆ Ba = r+ θ , 3 3 r r 1 2 where m = π a I a . c Then
∫
Φ Bb = Ba ⋅ dab
a
R=
z
z2 + b 2
. b
We can compute the flux over any surface that has the loop with radius b as its boundary (Problem 7.9, on this week’s homework); looks easiest if we take it over the sphere with radius R centered in the center of the small loop: 25 November 2002
Physics 217, Fall 2002
8
Coupled circular loops (continued) Φ Bb = ∫ Ba ⋅ dab = ∫ 2π m = R
θ0
∫ 0
2 m cos θ R3
R 2 sin θ dθ dφ
a
2θ 0
πm 2 cos θ sin θ dθ = R
∫
sin udu
0
πm πm πm 2θ 0 =− − cos u 0 = cos 2θ 0 R R R πm πm = − cos 2 θ 0 − sin 2 θ 0 R R π m π m z2 b 2 π m 2 2 2 2 = − 2 − 2 = 3 z + a − z + b R R R R R
(
=
( z2 + b )
25 November 2002
2 32
z
z2 + b 2
)
(
2π mb 2
R=
2π 2 = c
a2 b 2
Ia 3 2 2
( z2 + b )
)
b
.
Physics 217, Fall 2002
9
Coupled circular loops (continued) (c) Thus 2
Φ Ba Φ Bb 2π M= = = 2 cI b cI a c
a
2 2
(
b a z2 + b
)
2 32
. R=
Note that it’s much easier to get z2 + b 2 the fluxes -- and anything related to them, such as induced currents and fields -- by means of part a than part b. Replace 1 c 2 with µ0 4π for the MKS answer.
25 November 2002
Physics 217, Fall 2002
z
b
10
Transformers Back to our original loops. Suppose that the current in loop 1 varies with time. Then there is an EMF induced in loop 2: dI 1 1 dΦ B 2 = −M E2 = − c dt dt
.
dA 2
Thus
B1
r
I2 =
E2 M dI 1 =− R R dt
; I
dA 1
a time-varying current in one loop ends up inducing a timevarying current in the other, with relative size M. 25 November 2002
Physics 217, Fall 2002
11
Self inductance and back EMF What about one loop by itself? When one tries to change I, one changes the flux of the loop’s own B through itself, and dA ′ thus produces an opposing EMF. Φ B = ∫ B ⋅ da = v ∫ A ⋅ dA
B r dA
I dA ′ ⋅ dA = w ≡ cLI , where ∫∫ r c dA ′ ⋅ dA 1 L= 2 w ∫∫ r c Self inductance dA ′ ⋅ dA µ0 = in MKS . 4π w ∫∫ r
25 November 2002
Physics 217, Fall 2002
12
Self inductance and back EMF This opposing EMF, 1 dΦ B dI E =− = −L c dt dt
, Back EMF
is a form of inertia in electrical circuits. If one tries to change a current, a back EMF in induced. We will see soon that in the “equations of motion” for AC circuits, L plays the role that m plays in classical-mechanical equations. Work per unit charge done against a back EMF is just -E: dW dI P= = −EI = LI Compare: dt dt 1 2 1 2 = for capacitors. W CV . W = dW = L IdI = LI 2 2
∫
25 November 2002
∫
Physics 217, Fall 2002
13
Units Q r E] [ = = sec 2 cm -1 in cgs units; [L] = dI Q t dt t volt sec = ≡ henry in MKS units. amp 1 × 10 −11 sec 2 cm -1 1 henry ⇔ 9
Common values in the lab: L ≈ 1 mH - 1 µ H ≈ 10 −15 − 10 −18 sec2 cm -1 25 November 2002
Physics 217, Fall 2002
. 14
Calculation of inductance We have the two formulas, 1 dA 2 ⋅ dA 2 M= 2 r c
v∫
1
, L= 2 c
v∫
dA ′ ⋅ dA r
,
which turn out to be impractical in most cases for computing inductances. Far easier is the procedure resembling the calculation of capacitance: run a current, calculate the B produced, then the flux, and then use
Φ B1 = cMI 2
or Φ B = cLI
,
as we did in the example of the coupled circular loops. 25 November 2002
Physics 217, Fall 2002
15
Calculation of inductance (continued) For example: what’s the inductance per unit length of a very long solenoid with length A, radius a, and turns per unit length n? First run a current I through it, then the magnetic field inside is 4π B= nI c 2 2 2 4π a n A 4 π 2 ⇒ Φ B = BA = nI ⋅ nAπ a = I = cLI c c
⇒ L=
25 November 2002
4π 2 a2 n2 A c2
= µ0π a2 n2 A in MKS .
Physics 217, Fall 2002
16
25 November 2002
Physics 217, Fall 2002
dA 2
B1
r
I
dA 1
1
Mutual inductance Consider two coupled wire loops, arranged such that a current through one of them produces a magnetic field that has flux through the other one. For instance,
Φ B2 = ∫ B1 ⋅ da2 = ∫ ( — × A1 ) ⋅ da2 =v ∫ A1 ⋅ dA 2 But
A1 =
Φ B2
I1 c
I1 = c
dA 2
.
dA 1 , so r dA 2 ⋅ dA 2 ∝ I1 r
r
v∫
v∫
B1
;
I
dA 1
the proportionality factor is a constant that depends only upon the geometry of the loops. 25 November 2002
Physics 217, Fall 2002
2
Mutual inductance (continued) Now suppose we exchange the roles of the loops, and run a current through loop #2. We’d get the same result with just the 1s and 2s switched: I 2 dA 2 ⋅ dA 2 Φ B1 = ∝ I2 . r c
v∫
The proportionality factor is the same in the two cases. Define it as follows: Φ B1 = cMI 2 , Φ B2 = cMI 1 , where M=
1 c2
v∫
dA 2 ⋅ dA 2 r
µ0 = 4π
v∫
dA 2 ⋅ dA 2 in MKS r
is called the mutual inductance of the loops. 25 November 2002
Physics 217, Fall 2002
3
How mutual inductance can help with magnetic flux calculations This simple exercise has a profound conclusion: no matter what the shape of two loops, the flux through loop 2, resulting from current I in loop 1, is exactly the same as the flux through loop 1 due to current I in loop 2. This occasionally is useful in calculations of flux and related quantities: switching the currents and fluxes for the calculation, and then switching back, can save a lot of trouble. Consider Example 7.10: a short solenoid (length A, radius a, n1 turns per unit length) and a very long one (b, n2) are coaxial. Current I flows in the short one. What is the flux through the long one? 25 November 2002
Physics 217, Fall 2002
4
Coupled solenoids This would be difficult if calculated the way it’s stated, but easy if one instead runs the same current through the outer coil and calculates the flux through the inner: ΦB =
4π n2 I × n1 Aπ a2 c
.
a b
25 November 2002
Physics 217, Fall 2002
5
Coupled circular loops A small loop of wire (radius a) lies a distance z above the center of a large loop (radius b). The planes of the two loops are parallel, and perpendicular to the common axis. R= (a) Suppose current I flows in the big z2 + b 2 loop. Find the flux in the little loop. (b) Suppose current I flows in the little loop. Find the flux through the big loop. (c) What is the mutual inductance?
25 November 2002
Physics 217, Fall 2002
a
z
b
6
Coupled circular loops (continued) (a) We saw long time ago (Example 5.6, lecture notes for 4 November 2002) that Ib b2 2π B= zˆ , 3 2 c z2 + b 2 R=
(
)
∫
25 November 2002
(
z
z2 + b 2
so if the loop is small enough to consider B to be constant over its extent,
2π 2 Φ Ba = Bb ⋅ daa ≅ Bb Aa = c
a
b
b 2 a2 z2 + b
Ib 3 2 2
Physics 217, Fall 2002
)
.
7
Coupled circular loops (continued) (b) Supposing that the small loop is small enough to treat as a dipole, 2 m cos θ m sin θ ˆ ˆ Ba = r+ θ , 3 3 r r 1 2 where m = π a I a . c Then
∫
Φ Bb = Ba ⋅ dab
a
R=
z
z2 + b 2
. b
We can compute the flux over any surface that has the loop with radius b as its boundary (Problem 7.9, on this week’s homework); looks easiest if we take it over the sphere with radius R centered in the center of the small loop: 25 November 2002
Physics 217, Fall 2002
8
Coupled circular loops (continued) Φ Bb = ∫ Ba ⋅ dab = ∫ 2π m = R
θ0
∫ 0
2 m cos θ R3
R 2 sin θ dθ dφ
a
2θ 0
πm 2 cos θ sin θ dθ = R
∫
sin udu
0
πm πm πm 2θ 0 =− − cos u 0 = cos 2θ 0 R R R πm πm = − cos 2 θ 0 − sin 2 θ 0 R R π m π m z2 b 2 π m 2 2 2 2 = − 2 − 2 = 3 z + a − z + b R R R R R
(
=
( z2 + b )
25 November 2002
2 32
z
z2 + b 2
)
(
2π mb 2
R=
2π 2 = c
a2 b 2
Ia 3 2 2
( z2 + b )
)
b
.
Physics 217, Fall 2002
9
Coupled circular loops (continued) (c) Thus 2
Φ Ba Φ Bb 2π M= = = 2 cI b cI a c
a
2 2
(
b a z2 + b
)
2 32
. R=
Note that it’s much easier to get z2 + b 2 the fluxes -- and anything related to them, such as induced currents and fields -- by means of part a than part b. Replace 1 c 2 with µ0 4π for the MKS answer.
25 November 2002
Physics 217, Fall 2002
z
b
10
Transformers Back to our original loops. Suppose that the current in loop 1 varies with time. Then there is an EMF induced in loop 2: dI 1 1 dΦ B 2 = −M E2 = − c dt dt
.
dA 2
Thus
B1
r
I2 =
E2 M dI 1 =− R R dt
; I
dA 1
a time-varying current in one loop ends up inducing a timevarying current in the other, with relative size M. 25 November 2002
Physics 217, Fall 2002
11
Self inductance and back EMF What about one loop by itself? When one tries to change I, one changes the flux of the loop’s own B through itself, and dA ′ thus produces an opposing EMF. Φ B = ∫ B ⋅ da = v ∫ A ⋅ dA
B r dA
I dA ′ ⋅ dA = w ≡ cLI , where ∫∫ r c dA ′ ⋅ dA 1 L= 2 w ∫∫ r c Self inductance dA ′ ⋅ dA µ0 = in MKS . 4π w ∫∫ r
25 November 2002
Physics 217, Fall 2002
12
Self inductance and back EMF This opposing EMF, 1 dΦ B dI E =− = −L c dt dt
, Back EMF
is a form of inertia in electrical circuits. If one tries to change a current, a back EMF in induced. We will see soon that in the “equations of motion” for AC circuits, L plays the role that m plays in classical-mechanical equations. Work per unit charge done against a back EMF is just -E: dW dI P= = −EI = LI Compare: dt dt 1 2 1 2 = for capacitors. W CV . W = dW = L IdI = LI 2 2
∫
25 November 2002
∫
Physics 217, Fall 2002
13
Units Q r E] [ = = sec 2 cm -1 in cgs units; [L] = dI Q t dt t volt sec = ≡ henry in MKS units. amp 1 × 10 −11 sec 2 cm -1 1 henry ⇔ 9
Common values in the lab: L ≈ 1 mH - 1 µ H ≈ 10 −15 − 10 −18 sec2 cm -1 25 November 2002
Physics 217, Fall 2002
. 14
Calculation of inductance We have the two formulas, 1 dA 2 ⋅ dA 2 M= 2 r c
v∫
1
, L= 2 c
v∫
dA ′ ⋅ dA r
,
which turn out to be impractical in most cases for computing inductances. Far easier is the procedure resembling the calculation of capacitance: run a current, calculate the B produced, then the flux, and then use
Φ B1 = cMI 2
or Φ B = cLI
,
as we did in the example of the coupled circular loops. 25 November 2002
Physics 217, Fall 2002
15
Calculation of inductance (continued) For example: what’s the inductance per unit length of a very long solenoid with length A, radius a, and turns per unit length n? First run a current I through it, then the magnetic field inside is 4π B= nI c 2 2 2 4π a n A 4 π 2 ⇒ Φ B = BA = nI ⋅ nAπ a = I = cLI c c
⇒ L=
25 November 2002
4π 2 a2 n2 A c2
= µ0π a2 n2 A in MKS .
Physics 217, Fall 2002
16
