Today in Physics 122: pre-exam DC circuit and ...dmw/phy122/Lectures/Lect_30b.pdfToday in Physics...
Transcript of Today in Physics 122: pre-exam DC circuit and ...dmw/phy122/Lectures/Lect_30b.pdfToday in Physics...
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Today in Physics 122: pre-exam DC circuit and magnetostatics examples
A polygon current loop
Inductance of a toroid with circular loops
The torque you need to run a generator
Two (four?) time constants
11 November 2019 Physics 122, Fall 2019 1
R2 nπ
0I
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A polygon current loop
A wire is bent into the shape of a regular polygon with n sides whose vertices are a distance R from the center. If the wire carries a current I,
(a) determine the magnetic field at the center;
(b) if n is allowed to become very large show that the formula in part (a) reduces to that for a circular loop.
11 November 2019 Physics 122, Fall 2019 2
( ),n →∞
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A polygon current loop (continued)
(a) This is a superposition of nidentical Biot-Savart field law problems: B at some distance above the midpoint of a current segment.
Each segment has length
and its midpoint lies
from the center.
Need appropriate coordinate system and and to determine the distance from to the center.
11 November 2019 Physics 122, Fall 2019 3
RR
2 nπ
0x 0x
0y
( )02 2 sin ,x R nπ=
( )0 cosy R nπ=
n = 7 shown.
I
Id,Id
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A polygon current loop (continued)
The segments all produce B in the same direction: out of the page.
The contribution to B from each segment can be calculated separately. Follow recipe:
Point x along segment; point ythrough center of polygon, bisecting the segment.
11 November 2019 Physics 122, Fall 2019 4
0x− 0x
0y
x′
Idθ
′−r r
dB
x
y
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A polygon current loop (continued)
Identify an appropriate infinitesimal current element:
lying at
Get r – r’ for this element:
11 November 2019 Physics 122, Fall 2019 5
ˆ ,I Idx′=d x
.x′
2 2 20
02 2 2 2
0 0
ˆ ˆcos sin .ˆ ˆ
x y
yx
x y x y
θ θ
′ ′− = +
′− = +′
= − +′ ′+ +
r r
r r x y
x y0x− 0x
0y
x′
Idθ
′−r r
dB
x
y
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A polygon current loop (continued)
Now we have all the ingredients for the B-S law, and we integrate:
Substitute:
11 November 2019 Physics 122, Fall 2019 6
( )( )
( )
0
0
0
0
0 01 02 3 22 2
0
0 03 22 2
0
ˆ ˆ ˆ4 4
ˆ4
x
x
x
x
II dx x yx y
Iy dx
x y
µ µπ π
µπ
−
−
′ ′× − ′= = × − +′− ′ +
′=
′ +
∫ ∫
∫
d r rB x x yr r
z
( )2
0 03 22 2 2 2 00 0
21 cos sinsin cos2
y y x dxd dx
yx y x y
θ θθ θ θ′ ′
′= = − =′ + ′ +
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A polygon current loop (continued)
So
11 November 2019 Physics 122, Fall 2019 7
0 02 2 2 20 0 0 0
arccos arccos
.2 2 2 2
x x
x y x y
n n n n
θ
π π π π π π π ππ
= → − + +
= − → − − = − → +
0 0As ,x x x′ = − →
[ ]
2 30 0 0
1 3 202
20 02
0 0
sinˆ4 sin
ˆ ˆcos sin .4 2
n
n
nn
Iy y dy
I Iy y n
π π
π π
π ππ π
µ θθπ θ
µ µ πθπ π
+
−
+−
=
= − =
∫B z
z z
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A polygon current loop (continued)
Thus the total B from all n segments is
(b) If n is large, we may use the small-angle approximation,
whence
as expected (see lecture notes for 23 October 2019).
11 November 2019 Physics 122, Fall 2019 8
0
0ˆ sin .
2Iny n
µ ππ
=
B z
( )( )0
sin ,
cos ,
n n
y R n R
π π
π
≅
= ≅
0ˆ ,2
IR
µ=B z
http://www.pas.rochester.edu/%7Edmw/phy122/Lectures/Lect_22b.pdf
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Toroid with circular loops
(a) Show that the self-inductance L of a toroid of radius r0 containing N loops, each circular with diameter d, is
(b) Calculate L for d = 2.0 cm and r0 = 66 cm. Assume that the field inside the toroid is uniform, and that there are 550 loops in it.
11 November 2019 Physics 122, Fall 2019 9
2 20
00
if .8N d
L r dr
µ≈
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Toroid with circular loops (continued)
(a) We to hypothesize a current I for the toroid; to calculate B inside the toroid; and to calculate in a typical loop, in order to use
The B calculation can use Ampère’s law, with a circular path concentric with the toroid:
11 November 2019 Physics 122, Fall 2019 10
( )
0 encl
0
0
2
ˆ2
I
B r NINIr
µ
π µµπ
=
=
= −
∫ B d
B
φ
r0
.BL N I= Φ
BΦ
x
y
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Toroid with circular loops (continued)
If then B doesn’t vary by much from the inner point on the loop to the outer point; we can take
which simplifies the flux calculation:
Thus the self-inductance becomes
as advertised. 11 November 2019 Physics 122, Fall 2019 11
2 20
0, q.e.d.
8B N dNL
I rµΦ
= ≅
0d r
0
0
ˆ2
NIr
µπ
≅ −B φ ,
20
0loop
.2 4B
NI dr
µ ππ
Φ = ≅∫ B dAr0
x
y
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Toroid with circular loops (continued)
(b) The arithmetic:
11 November 2019 Physics 122, Fall 2019 12
52.9 10 henry.L −= ×
r0
x
y
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The torque you need to run a generator
A generator is rated at 16 kW, 250 V and 64 A when it rotates at 1000 rpm. The resistance of the armature windings is 0.40 Ω. Assume that the magnitude of the magnetic field in which the armature rotates is constant.
(a) Calculate the “no load” voltage at 1000 rpm – no load meaning that no circuit is hooked up to the generator.
(b) Calculate the full-load voltage (i.e. at 64 A) when the generator is run at 750 rpm.
(c) Calculate the torque necessary to produce the motor’s rated performance at 1000 rpm.
11 November 2019 Physics 122, Fall 2019 13
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The torque you need to run a generator (continued)
(a) The equivalent circuit of the fully-loaded generator is at right; the loop equation gives us
With no load (no current drawn), the output voltage would be just
11 November 2019 Physics 122, Fall 2019 14
0 sin tω=
0.4r = Ω 0 64 ampsI =
0 250 voltsV = R
0 0 0
0 0 0
0280 volts .
I r VV I r
− − == + =
0 280 V.=
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The torque you need to run a generator (continued)
(b) If there are N windings in the motor, and each has area A, then the EMF is
So
11 November 2019 Physics 122, Fall 2019 15
( )
0
0
cos
sin sin ;.
Bd d NAB tdt dt
NAB t tNAB
ω
ω ω ωω
Φ= − = −
= ==
( ) ( )
( ) ( )
20 2 0 1
1
22 0 1
1
750 250 V 190 V.1000
RVr R
ωω ω
ωω
ω ωω
=
= = =+
0 sin tω=
0.4r = Ω 0 64 ampsI =
0 250 voltsV = R
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The torque you need to run a generator (continued)
(c) Again, If there are N windings in the armature and each winding has area A, then the magnetic field exerts a torque on the armature:
In generators this is called counter torque; it’s the generator version of the DC motor’s back emf. You need to exert an equal and opposite torque to maintain the armature’s rotation.
Note that this requires continuous exertion of mechanical power, as you may have guessed from conservation of energy:
11 November 2019 Physics 122, Fall 2019 16
( )mech elec .
dW d dtdWP NIAB NIB I I Pdt
τ θ τω
τω ω ω
= =
= = = = = =
( )( )( ) ( )1280V 64A 1000 2 60 rad sec 0.05 N m .
B NIA B Iτ µ ω
π −= = =
= × × =
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Two time constants
Initially neither of the circuits at right has any current. Then switch S is closed. After a long time, it is opened again. Find the time constants according to which the current changes, for the opening and closing of the switches. Are they different?
11 November 2019 Physics 122, Fall 2019 17
1R
1R
2R
2R
S
S
L
C
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Two time constants (continued)
This is a pair of Kirchhoff-rule problems. We’re after the current in Land C.
First the inductor: the node equation, and the loop equations for the two small loops, are
Solve the node equation for I1:
11 November 2019 Physics 122, Fall 2019 18
1R
1R
2R
2R
S
S
L
C
1I
2I3I
1 2 3
1 1 2 2
32 2
00
0
I I II R I R
dII R L
dt
− − =− − =
− =
1 2 3I I I= +
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Two time constants (continued)
Substitute the result into the first loop equation and solve for I2:
Substitute this result into the second loop equation and solve for
11 November 2019 Physics 122, Fall 2019 19
1R
1R
2R
2R
S
S
L
C
1I
2I3I( )
( )2 2 1 2 3 1
2 1 2 3 1
3 12
1 2
I R I R I I RI R R I R
I RI
R R
= − = − +
+ = −
−=
+
3 :dI dt
( )
3 3 12 2 2
1 2
1 23
1 2 1
dI I RI R Rdt L R R L
R R IL R R R
−= =
+
= − −
+
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Two time constants (continued)
Separate variables, with current on the left and time on the right:
and identify the time constant as the reciprocal of the factor between the minus and dt:
11 November 2019 Physics 122, Fall 2019 20
1R
1R
2R
2R
S
S
L
C
1I
2I3I
( )3 1 2
3 1 1 2
dI R R dtI R L R R
= −− +
( )1 21
1 2L
L R RR R
τ+
=
-
Two time constants (continued)
Now for the second circuit: the Kirchhoff rules give us
Follow the same steps as before, using Q/C where we had
11 November 2019 Physics 122, Fall 2019 21
1R
1R
2R
2R
S
S
L
C
1I
2I3I
1 2 3
1 1 2 2
2 2
00
0
I I II R I R
QI RC
− − =− − =
− =
3 :LdI dt
3 12 2 2
1 2
21
1 2
I RQ I R RC R R
R C dQQ RR R dt
−= =
+
= − − +
-
Two time constants (continued)
And rearrange:
once again to identify the time constant as the reciprocal of the factor between the minus and dt:
11 November 2019 Physics 122, Fall 2019 22
1R
1R
2R
2R
S
S
L
C
1I
2I3I
2
1 2
1 2 2
1 2 1 2
1 22 1 2
1 2
R C dQQR R dtR R C R CdQR R dt R RR RdQ dtR C R R CQ
R R
= − − +
= − ++ ++
= −−
+
1 21
1 2C
R R CR R
τ =+
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Two time constants (continued)
Now open the switches. No more current can flow through the first resistor, so we get the simple RL and RC circuits, for which the time constants are L/R and RC.
So, yes, they are different.
11 November 2019 Physics 122, Fall 2019 23
1R
1R
2R
2R
S
S
L
C2I
2R2I
Today in Physics 122: pre-exam DC circuit and �magnetostatics examplesA polygon current loopA polygon current loop (continued)A polygon current loop (continued)A polygon current loop (continued)A polygon current loop (continued)A polygon current loop (continued)A polygon current loop (continued)Toroid with circular loopsToroid with circular loops (continued)Toroid with circular loops (continued)Toroid with circular loops (continued)The torque you need to run a generatorThe torque you need to run a generator (continued)The torque you need to run a generator (continued)The torque you need to run a generator (continued)Two time constantsTwo time constants (continued)Two time constants (continued)Two time constants (continued)Two time constants (continued)Two time constants (continued)Two time constants (continued)