Assignment 10 with Solution - Engineering Physics 2 | PHYS 216 | Assignments Physics

Physics 216, Assignment 10 solutions, Fall 2008

29.2. IDENTIFY:

ΔΦ

=Δ

E. cos

BBA

Φ= .

is the flux through each turn of the coil.

SET UP: . .

=°f90

=°

EXECUTE: (a) The total flux through the coil is

542 8

,i cos0 (6.0 10 T)(12 10 m )(1) 7.2 10 Wb.

BBA −− −

Φ= = × × = ×°

7.2 10 Wb) 1.44 10 Wb

−−

× = × ,f cos90 0

BBA

,i (200)(

NΦ=

==°.

(b) 54

1.44 10 Wb 3.6 10 V 0.36 mV

0.040 s

−−

Φ− Φ ×

==×=

=E.

EVALUATE: The average induced emf depends on how rapidly the flux changes.

29.4. IDENTIFY and SET UP: Apply the result derived in Exercise 29.3: /.QNBAR

In the present exercise the flux

changes from its maximum value of B

AΦ= to zero, so this equation applies. R is the total resistance so here

60.0 45.0 105.0 .

R=Ω+Ω= Ω

EXECUTE: 5

(3.56 10 C)(105.0 )

says 0.0973 T.

120(3.20 10 m )

NBA QR

RNA

−

×Ω

=== =

EVALUATE: A field of this magnitude is easily produced.

29.17. IDENTIFY and SET UP: Apply Lenz's law, in the form that states that the flux of the induced current tends to

oppose the change in flux.

EXECUTE: (a) With the switch closed the magnetic field of coil A is to the right at the location of coil B. When

the switch is opened the magnetic field of coil A goes away. Hence by Lenz's law the field of the current induced

in coil B is to the right, to oppose the decrease in the flux in this direction. To produce magnetic field that is to the

right the current in the circuit with coil B must flow through the resistor in the direction a to b.

(b) With the switch closed the magnetic field of coil A is to the right at the location of coil B. This field is stronger

at points closer to coil A so when coil B is brought closer the flux through coil B increases. By Lenz's law the field

of the induced current in coil B is to the left, to oppose the increase in flux to the right. To produce magnetic field

that is to the left the current in the circuit with coil B must flow through the resistor in the direction b to a.

circuit that includes coil A increases when R is decreased and the magnetic field of coil A increases when the

current through the coil increases. By Lenz's law the field of the induced current in coil B is to the left, to oppose

the increase in flux to the right. To produce magnetic field that is to the left the current in the circuit with coil B

must flow through the resistor in the direction b to a.

EVALUATE: In parts (b) and (c) the change in the circuit causes the flux through circuit B to increase and in part

(a) it causes the flux to decrease. Therefore, the direction of the induced current is the same in parts (b) and (c) and

opposite in part (a).

29.18. IDENTIFY: Apply Lenz’s law.

SET UP: The field of the induced current is directed to oppose the change in flux in the primary circuit.

EXECUTE: (a) The magnetic field in A is to the left and is increasing. The flux is increasing so the field due to the

induced current in B is to the right. To produce magnetic field to the right, the induced current flows through R

from right to left.

(b) The magnetic field in A is to the right and is decreasing. The flux is decreasing so the field due to the induced

current in B is to the right. To produce magnetic field to the right the induced current flows through R from right to

left.

current in B is to the left. To produce magnetic field to the left the induced current flows through R from left to right.

EVALUATE: The direction of the induced current depends on the direction of the external magnetic field and

whether the flux due to this field is increasing or decreasing.

29.30. IDENTIFY: Use Eq.(29.10) to calculate the induced electric field E and use this E in Eq.(29.9) to calculate E

between two points.

(a) SET UP: Because of the axial symmetry and the absence of any electric charge, the field lines are concentric

circles.

29-1

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Physics 216, Assignment 10 solutions, Fall 2008

29.2. IDENTIFY: B

E. Φ B = BA cos φ. Φ B is the flux through each turn of the coil.

S ET U P: φ =i 0 °. φ =f 90 °.

Φ = °= × −^ × −^ = × −

N Φ B ,i = (200)( Φ B ,f = BA cos90 ° = 0.

N N

E =.

(3.56 10 C)(105.0 )

NBA QR

Q B

R NA

× Ω

×

E

G

of E

G

E XECUTE: Use the sign convention for Faraday’s law to deduce this direction. Let A be into the paper. Then

G

clockwise. Thus E

G

G G

E ⋅ d l = E (2 π r )

G G

B

π = π = = = × −

(c) The induced emf has magnitude = E ⋅ d l = E (2 π r ) = (1.75 × 10 −^3 V/m)(2 π)(0.100 m) = 1.100 × 10 −^3 V.

G G

I

R

E

(d) Points a and b are separated by a distance around the ring of π r so

E= E ( π r ) = (1.75 × 10 −^3 V/m)( π)(0.100 m) = 5.50 × 10 −^4 V

(e) The ends are separated by a distance around the ring of 2 π r so E= 1.10 × 10 −^3 Vas calculated in part (c).

29.31. IDENTIFY: Apply Eq.(29.1) with Φ B = μ 0 niA.

S ET U P: A = π r^2 , where r = 0.0110 m. In Eq.(29.11), r = 0.0350 m.

E XECUTE: B ( ) ( 0 ) 0

E = = = =

and E = E (2 π r ).Therefore,

di E r

dt nA

×

EVALUATE: Outside the solenoid the induced electric field decreases with increasing distance from the axis of

the solenoid.

29.36. IDENTIFY and S ET U P: Eqs.(29.13) and (29.14) show that i C = i Dand also relate to the rate of change of the

electric field flux between the plates. Use this to calculate and apply the generalized form of Ampere’s law

(Eq.29.15) to calculate B.

i D

dE dt /

0.280 A 0.280 A

, so 55.7 A/m

(0.0400 m)

i i

i i j

A A π r π

55.7 A/m

so 6.29 10 V/m s

8.854 10 C / N m

dE dE j

j

dt dt −

= = = = ×

× ⋅

P

P

(c) S ET UP: Apply Ampere’s law B ⋅ d l = μ 0 ( i C + i D )encl

G G

E XECUTE: Wire A : v

G

G

G G

Wire C : v is perpendicular to B. The mponent of v

G

E = (0.35 0 m/s)(cos 45 )(0.120° )(0.500 m) =0..

G