Matter & Interactions Chapter18 Solutions Q12: Solutionphysics.drexel.edu/~vogeley/Phys115/Solutions/hw2_sols.pdf · Matter & Interactions Chapter18 Solutions Q15: Solution: (a)Nocurrentwillﬂowthroughthewire(s)

Matter & Interactions Chapter 18 Solutions

Q12:Solution:

To quote from pg. 769 of the textbook, “It is important to keep in mind that although the units of emf are volts, the emfis not a potential difference. Potential difference is a path integral of the electric field made by real charges. The emf isthe energy input per unit charge and might in principle be gravitational or nuclear in nature.” In the case of a battery,the emf of the battery is produced by chemical energy input per unit charge.

Q13:Solution:

The battery maintains positive surface charge on the positive terminal and negative surface charge on the negativeterminal. The electric field within the battery points from the positive terminal toward the negative terminal. Withinthe wires, the electric field also points away from the positive terminal and toward the negative terminal. However, in asketch of the circuit, the electric field just inside the battery will be opposite the electric field within the wire just outsidethe battery.

Q14:Solution:

(a) According to conservation of charge (the node rule), i1

= i2. Thus,

i1

= i2

nAu1E

1= nAu

2E

2

u1E

1= u

2E

2

E2

=

(u1

u2

)E

1

=1

3E

1

(b) Neglect the potential difference (energy loss) across the connecting wires. Apply the loop rule to the circuit withthe single bulb (bulb 1, which we will call 0 to distinguish this case from the previous case in part (a)).

emf − E0L = 0

The current is

i0

= nAu1E

0

= nAu1

(emf

L

)

Now apply the loop rule to the circuit with two bulbs.

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emf − E1L− E

2L = 0

emf − E1L−

(1

3

)E

1L = 0

emf − 4

3E

1L = 0

E1

=3

4

emf

L

The current through bulb 1 is

i1

= nAu1E

1

= nAu1

(3

4

emf

L

)

But we know that

i0

= nAu1

(emf

L

)

So, substitution gives

i1

=3

4i0

Evidently replacing Bulb 2 with a wire increased the current through the battery by a factor of 4/3. Or, to say itin another way, if you originally have a battery and bulb 1 and you connect bulb 2 in series with bulb 1, the currentthrough the battery and bulbs will be 3/4 of the original current.

(c)

iCu

= ifilament

nCuA

CuuCuE

Cu= n

filA

filufilE

fil

The cross-sectional area of a wire is at least 100 times larger than the area of a filament. The mobility of copper isabout 10 times larger than tungsten. Thus, the electric field in the copper wire is at least 1000 times smaller thanin the tungsten.

The electric fields in all the copper wires are the same because i = nAuE and i (and n, A, and u) is the same inall the copper wires. Thus, E must also be the same.

(d) The graph is shown in the figure below. Gray lines are for the circuit with two bulbs (part a).

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Q15:Solution:

(a) No current will flow through the wire(s). The reason is that if you apply the loop rule to the entire circuit, you get

emf1 − emf2 −∆Vwire = 0

where ∆Vwire is the potential difference across the entire long wire, measured from the negative terminal of battery1 to the negative terminal of battery 2. Since the batteries are identical emf1 = emf2, so ∆Vwire = 0. As a result,E = 0 and i = nAuE = 0 everywhere inside the wire(s). This means that the surface charge density must beuniform along the wires. There is no gradient (i.e. change) in the surface charge density. A sketch is shown in thefigure below.

____

____

____

____

____

____

____

____

____

____

_ _ _ _

_ _ _ _

++++

++++

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(b) The electric field is zero at all locations inside the wire.

(c) The brightness of the bulb in this case is zero. There is no current through the bulb. If one of the batteries wasnot connected backwards, then there would be current through the bulb and the bulb would light.

(d) It is worth doing the experiment to “see” the results for yourself.

Q16:Solution:

Note that the battery remains the same in all experiments, and according to the Loop Rule, emf = ∆V where ∆V is thepotential difference across the wire. Since ∆V = EL, then emf = EL where L is the length of the wire. This can be auseful relationship. Answers are shown in Table ??.

Experiment Effect on current Parameter that changedDouble L i changes by a factor of 1/2 E; E changes by a factor of 1/2

because L increased by 2 and i ∝A.

Double A i changes by a factor of 2 A; A changes by a factor of 1/2and i ∝ A.

Double L i changes by a factor of 1/2 E; Two bulbs in series is like hav-ing a single filament that is twiceas long. Doubling L changes Eby 1/2 and i ∝ E.

Double emf i changes by factor of 2 E; 2emf results in 2E which re-sults in 2i since i ∝ E.

Q17:Solution:

The resistor (thin wire) and thick connecting wires are in series. Applying conservation of charge (i.e. the Current NodeRule) to the resistor and wires leads to the conclusion that the electron current through the thin wire must equal theelectron current through the thick wire. Use i = neAuE and the fact that n and u are properties of the material in orderto compare the quantities given in Table ??.

resistor <, =, or > thick wiresiR = iwnR = nwAR < AwuR = uwER > EwvR > vw

Q18:Solution:

(a) The electric field is uniform throughout the wire and points away from the higher potential (i.e. positive) terminalof the battery and toward the lower potential (i.e. negative) terminal of the battery. A sketch of ~E is shown in thefigure below.

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i∆T = NI

|q|∆T = N

∆T =N |q|I

≈( 1

2 mol)(

6× 1023

ions/mol)(

1.6× 10−19

C/ion)

0.3 C/s

≈ 1.6× 105

s ≈ 44.4 h

P33:Solution:

True: 1False: 2, 3, 4, 5

In steady state, the currents at B and D must be equal so iB≈ i

D≈ 6× 10

18

e/s.

Use the properties of the thick wire to calculate the electric field inside it.

iB

= nBA

BuB

∣∣∣~EB

∣∣∣∣∣∣~EB

∣∣∣ =iB

nBA

BuB

≈ 6× 1018

electrons/s(4× 1028 electrons/m3

) (π(0.55× 10−3 m

)2)(6× 10−4 m/s

N/C

)≈ 0.26 N/C

P34:Solution:

(a) Electrons move to the right. Their magnetic field must point toward the top of the page.

(b) The current divides equally between bulbs 2 and 3 since they are identical. Thus the current through each of themis 1.5× 10

18

e/s.

(c)(B

2= B

3

)< B

1

(d)

i = nAu∣∣∣~E∣∣∣∣∣∣~E∣∣∣ =

i

nAu

≈ 3× 1018

electrons/s(6.3× 1028 electrons/m3

) (1× 10−8 m2

) (1.2× 10−4 m/s

N/C

)≈ 39.7 N/C

The electric field points to the left.

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P35:Solution:

(a) The needle deflects downward.

(b) The needle deflects 13◦ upward.

(c) ∣∣∣~E1

∣∣∣ =i

nAu

≈ 1.5× 1018

electrons/s(6.3× 1028 electrons/m3

) (1× 10−8 m2

) (1.2× 10−4 m/s

N/C

)≈ 19.8 V/m

~E1points to the left. In bulb 3, the electric field is to the right.

P36:Solution:

(a) Treat the battery as if it is a capacitor made of two oppositely charged discs separated a distance 4.5 cm. Tomake a rough estimation, assume that the electric field is nearly uniform between the plates and given by that ofa capacitor.

E ≈ Q/A

ε0

If the electric field is uniform, then the potential difference is ∆V = EL where L is the distance between theterminals and ∆V is equal to the emf of the battery.

∆V

L=

Q/A

ε0

Q =ε0A∆V

L

=(8.85× 10−12 C2/(N · m2))π(0.01 m)2(1.7 V)

0.045 m

= 1.05× 10−13 C

This is a very small amount of charge. It is also a very rough approximation because the equation for the electricfield within a capacitor was derived for the case of small separation between the discs (compared to the radius ofthe discs).

(b) The charge on a piece of tape is on the order of nanocoulombs (10−9 C). The charge on the terminal of a batteryis about 10,000 times smaller than the charge on a piece of tape. Thus, the coulomb force between the terminalof the battery and the tape would be 10,000 times smaller than the coulomb force between two identical pieces oftape. This is too small to be noticeable.

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Use this constant to find i1in this new circuit.

i1

= nA1u

2

3

emf

L

=2

3(9× 1017 s−1)

= 6× 1017 s−1

It makes sense that this current is greater than with two identical bulbs because it is easier to push current througha thick filament than a thin filament.

P50:Solution:

(a) Apply the loop rule. Call the thin wire 1 and the thick wire 2.

2emf − E1L

1− E

2L

2= 0

Apply the node rule.

i1

= i2

nA1uE

1= nA

2uE

2

A1E

1= A

2E

2

E1

=A

2

A1

E2

=(0.35)2

(0.25)2E

2

= 1.96 E2

≈ 2E2

Substitute into the loop equation.

2emf − 2E2L

1− E

2L

2= 0

2emf − E2(2L

1+ L

2) = 0

E2

=2emf

2L1

+ L2

=2(1.5 V)

2(0.5 m) + 0.15 m

= 2.6V

m

So, E1

= 2E2

= 5.2 Vm .

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(b) v̄ = uE, so calculate drift speed and ∆t for each wire seperately since v̄ is different for each wire.

v̄1 = (7× 10−5 m/s

V/m)(5.2

V

m)

= 3.64× 10−4 m/s

∆t1

=L

1

v̄1

=0.5 m

3.64× 10−4 m/s

= 1374 s

For wire 2,

v̄2 = uE2

= 1.82× 10−4 m/s

∆t2

=0.15 m

1.82× 10−4 m/s

= 824 s

∆t = ∆t1

+ ∆t2

= 1374 s + 824 s

≈ 2200 s

≈ (2200 s)

(1 mm

60 s

)= 36.6 min

(c) The minimum time required to reach steady state is given by the speed of light. Using L = L1

+ L2

= 0.65 m,

c =L

∆t

∆t =L

c

=0.65 m

3× 108 m/s

≈ 2× 10−9 s

This is 2 ns, which is much smaller than the time interval for a mobile electron to travel around the circuit.

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(d) The electron current is

i1

= nA1uE

1

= (9× 1028 m−3)π

(0.25× 10−3 m

2

)2

(7× 10−5 m/s

V/m)(5.2

V

m)

= 1.6× 1018 s−1

Thus, 1.6× 1018 electrons pass from the thin wire to the thick wire every second.

P51:Solution:

(a) The loop rule gives

emf − EL = 0

E =emf

L

=1.7 V

0.75 m

= 2.27V

m

(b) The electric field remains the same. It only depends on ∆V and L. However, the current will change.

(c) i = nAuE. i ∝ u so 4u gives 4i. As a result, (3) is true.

P52:Solution:

(a) Conventional current flows from the positive terminal of one battery to the negative terminal of the other battery.For the upper compass in the first circuit, conventional current flows south which creates a magnetic field beneaththe wire that is east. Thus, the compass points NE. For the lower compass in the first circuit, conventional currentflows north which creates a magnetic field beneath the wire that is west. Thus, the compass points NW. For thecompass with bulb B in the second circuit, the current flows south so the compass is deflected to the NE. A sketchis shown in the figure below. For clarity, angles are shown greater than what they would appear on an actualcompass.

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Note that this is the same as for bulb A in the one-bulb circuit. Thus, the compass next to bulb B will also deflect5◦.Applying the node rule to bulbs B and C leads to the conclusion that the current entering the negative terminalof the battery is equal to i = iB + iC = 2iB . Since the current through the batteries is twice the current throughbulb B, then current will create a magnetic field that is twice the magnitude of the magnetic field created by thecurrent through bulb B.The compass deflection is given by

tan θ =BwireBearth

Bwire,B = tan(5◦)(2× 10−5 T)

= 1.75× 10−6 T

Now, multiply this magnetic field by 2 and solve for θ. This gives

tan θ =BwireBearth

=3.5× 10−6 T

2× 10−5 Tθ = 9.9◦

Note that for small angles tan θ ≈ θ, so twice the magnetic field results in approximately twice the deflection angle.

(c) In general, there is a surface charge gradient that changes from high positive surface charge density at the positiveterminal of the battery to high negative surface charge density at the negative terminal of the battery. And at themidpoint of the circuit, the surface charge density will be zero because it is the point of transition from positivesurface charge density to negative surface charge density.Because the electric field in the thin filament of the bulb is very large compared to the connecting wires, then thegradient (i.e. change) in surface charge density along the filament will be very large. Because the electric field inthe thick wire is smaller, then the gradient (i.e. change) in surface charge density along the thick wire will be verysmall.For the one-bulb circuit, sketch a few positive charges at the positive side of bulb A and a few negative charges atthe negative side of bulb A. Then, sketch a a couple more positive surface charges at the positive terminal of thebattery and a couple more negative surface charges at the negative terminal of the other battery. There will behigh density positive and negative surface charge the other terminals of the two batteries where they are connectedby a single wire. For the two bulb circuit, make a similar sketch.

(d) Applying the loop rule to the one-bulb circuit gives

2emf − EAL = 0

EA =2emf

L

=2(1.5 V)

0.004 m= 750 V/m

Since the potential difference across bulb B and bulb C is also 3.0 V and since they are identical bulbs, thenEB = EC = EA = 750 V/m. Since bulbs B and C have the same current and potential difference across them asbulb A, then they will have the same brightness.

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A sketch of the compasses is shown in the figure below. The deflection angles in the sketch are exaggerated for greaterclarity. Note that conventional current flows from the + terminal of one battery to the – terminal of the second battery.As a result, the compass near the + terminal deflects NW since ~Bwire is west, and the compass near the – terminaldeflects NE since ~Bwire is east.

North 4 deg

15 deg

19 deg

19 deg

P54:Solution:

Since this system takes a very long time to come to static equilibrium, we can consider it to be in a quasi steady state.That is, at any instant we can assume electron current i is the same at all locations in the wire. We are interested in theelectron current in the wire a short time ∆t after the connection has been made–long enough for the rearrangement ofsurface charge to have occurred (this takes only nanoseconds), but short enough that the charge on the two spheres haschanged very little.Since i = nAuE, and we know n, A, and u, we need to find the electric field E in the wire. There are three sourcesof charge contributing to E in the wire: the two charged spheres, and the charge on the surface of the wire. Althoughthere is not a lot of charge on the wire, this charge is very close to the wire, and its contribution to E inside the wire issignificant. Since we donÕt know the exact amount and distribution of charge on the wire, we can’t calculate E directlyfrom Coulomb’s law.Therefore, we start (as usual) from a fundamental principle: ∆V = 0 for a round trip path. We choose a path that goesthrough the wire, where we want to know E, and also through the air, as shown in the figure below.

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Consider the part of the path indicated by a dotted line. Along the sections of the path inside the spheres, the electricfield is zero, and hence ∆V is zero.

Along the section of the path in the air, if we assume that (1) we are sufficiently far from the wire that the small amountof charge on the wire contributes a negligible electric field, and (2) the charged spheres are sufficiently far from each otherthat the electric field of one is negligible near the other, then we find that along the dotted line portion of the path:

∆Vair = VB − VA≈ 1

4πεo

(−qr− Q

R)

So for a round trip path from A back to A, we have:

∆V = 0

∆Vwire + ∆Vair = ∆Vwire +1

4πεo

(−qr− Q

R)

Since the system is in a quasi steady state, E must be uniform in the wire, so ∆Vwire = EL (it is positive, since we aregoing from B to A, traveling opposite to the direction of E in the wire). Thus,

E =

14πε

o

( qr + QR )

L

and

i = nAuE

= nAu

14πε

o

( qr + QR )

L

Therefore, a number of electrons equal to nAu1

4πεo

( qr+QR )

L ∆t leave the small sphere in a time ∆t.

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Matter & Interactions Chapter18 Solutions Q12: Solutionphysics.drexel.edu/~vogeley/Phys115/Solutions/hw2_sols.pdf · Matter & Interactions Chapter18 Solutions Q15: Solution: (a)Nocurrentwillﬂowthroughthewire(s)