Today in Physics 122: pre-exam DC circuit and ...dmw/phy122/Lectures/Lect_30b.pdfToday in Physics 122: pre-exam DC circuit and magnetostatics examples A polygon current loop Inductance

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

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.


( ),n →∞

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.


RR

2 nπ

0x 0x

0y

( )02 2 sin ,x R nπ=

( )0 cosy R nπ=

n = 7 shown.

I

Id,Id


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.


0x− 0x

0y

x′

Idθ

′−r r

dB

x

y


Identify an appropriate infinitesimal current element:

lying at

Get r – r’ for this element:


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


Now we have all the ingredients for the B-S law, and we integrate:

Substitute:


( )( )

( )

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′ ′

′= = − =′ + ′ +


So


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


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).


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

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.


2 20

00

if .8N d

L r dr

µ≈

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:


( )

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


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


(b) The arithmetic:


52.9 10 henry.L −= ×

r0

x

y

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.


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


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.=


(b) If there are N windings in the motor, and each has area A, then the EMF is

So


( )

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


(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:


( )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 −= = =

= × × =

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?


1R

1R

2R

2R

S

S

L

C

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:


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= +


Substitute the result into the first loop equation and solve for I2:

Substitute this result into the second loop equation and solve for


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

−= =

+

= − −

+


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:


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+

=


Now for the second circuit: the Kirchhoff rules give us

Follow the same steps as before, using Q/C where we had


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

−= =

+

= − − +


And rearrange:

once again to identify the time constant as the reciprocal of the factor between the minus and dt:


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 =+


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.


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)

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Today in Physics 122: pre-exam DC circuit and ...dmw/phy122/Lectures/Lect_30b.pdfToday in Physics 122: pre-exam DC circuit and magnetostatics examples A polygon current loop Inductance