H.W. 29/30

Seite 1

A Magnet and a Coil

When a magnet is plunged into a coil at speed v, as shown in the figure, a voltage is induced in the coil and a current flows in the circuit.

If the speed of the magnet is doubled, the induced voltage is ________ .- TWICE AS GREAT

The same magnet is plunged into a coil that has twice the number of turns as before. The magnet is shown before it enters the coil in the figure. (Figure 2)If the speed of the magnet is again v, the induced current in the coil is _______ .- UNCHANGED

(By increasing the number of turns in the coil, the induced emf increases, but so does the resistance of the coil. Since those two quantities increase by the same factor, their ratio remains constant, and the induced current in the circuit is unchanged.)

Seite 2

Conceptual Induction

A loop of wire is initially held above a short solenoid. A constant counterclockwise (as viewed from above) current I passes through the turns of the solenoid.

The loop of wire is steadily lowered, eventually "encircling" the solenoid. (Figure 1)

Throughout this problem, when you answer questions about the direction of current, assume that you are viewing the wire loop from above, looking downward.

What is the direction of the induced current in the loop when the loop is above the solenoid, moving downward?- CLOCKWISE

What is the direction of the induced current at the instant that the loop is at the midpoint of the solenoid and still moving downward?- NO CURRENT

What is the direction of the induced current when the loop is below the solenoid and moving downward?- COUNTERCLOCKWISE

Seite 3

Faraday's Law and Induced Emf

Learning Goal:

To understand the terms in Faraday's law and to be able to identify the magnitude and direction of induced emf.

Faraday's law states that induced emf is directly proportional to the time rate of change of magnetic flux. Mathematically, it can be written as

E=−ΔΦBΔt,

where E is the emf induced in a closed loop, and

ΔΦBΔt

is the rate of change of the magnetic flux through a surface bounded by the loop. For uniform magnetic fields the magnetic flux is given by ΦB=B⃗ ⋅A⃗ =BAcos(θ), where θ is the angle between the magnetic field B⃗ and the normal to the surface of area A.

To find the direction of the induced emf, one can use Lenz's law:

The induced current's magnetic field opposes the change in the magnetic flux that induced the current.

For example, if the magnetic flux through a loop increases, the induced magnetic field is directed opposite to the "parent" magnetic field, thus countering the increase in flux. If the flux decreases, the induced current's magnetic field has the same direction as the parent magnetic field, thus countering the decrease in flux.

Recall that to relate the direction of the electric current and its magnetic field, you can use the right-hand rule: When the fingers on your right hand are curled in the direction of the current in a loop, your thumb gives the direction of the magnetic field generated by this current.

In this problem, we will consider a rectangular loop of wire with sides x and yplaced in a region where a uniform magnetic field B⃗ exists as shown in (Figure 1). The resistance of the loop is R. For the purposes of this problem we will assume that the area vector A⃗ points out of the page.

Initially, the field is perpendicular to the plane of the loop and is directed out of the page. The loop can rotate about either the vertical or horizontal axis, passing through the midpoints of the opposite sides, as shown.

Which of the following changes would induce an electromotive force (emf) in the loop? When you consider each option, assume that no other changes occur.

Find the flux ΦB through the loop.

If the magnetic field steadily decreases from B to zero during a time interval t, what is the magnitude E of the induced emf?

Express your answer in terms of x, y, B, and t.

If the magnetic field steadily decreases from B to zero during a time interval t, what is the magnitude I of the induced current?

Express your answer in terms of x, y, B, t, and the resistance R of the wire.

If the magnetic field steadily decreases from B to zero during a time interval t, what is the direction of the induced current?

Which of the following changes would result in a clockwise emf in the loop? When you consider each option, assume that no other changes occur.

Seite 4

Problem 29.18

The magnetic flux through each loop of a 75-loop coil is given by (8.8t−0.51t^3)×10^−2(T⋅m^2), where the time t is in seconds.

Determine the emf as a function of time.

Express all the numbers using two significant figures. Express your answer in terms of the variable t.

The induced emf in the coil is

(Faraday's Law)- Emf= -N(dBA/dt)

{N=number of turns

{BA= magnetic flux- B(external magnetic field) A(area of coil)

N-75

BA= (8.8t-0.51t^3)*10^-2

d/dt(BA)= (8.8-0.51(3t^2))10^-2

(-6.6+1.15t^2)V

What is E at t = 1.1 s ?

Express your answer using two significant figures.

At t=1.1s, the induced emf is

(-6.6+1.14(1.1)^2)

= -5.2V

What is E at 3.5 s ?

Express your answer using two significant figures

At t=3.5s, the induced emf is

(-6.6+1.14(3.5)^2)

=7.5V

Seite 5

Problem 29.26

Determine the emf induced in the square loop in the figure if the loop stays at rest and the current in the straight wire is given by I(t)=(15.0A)sin(2500t) where t is in seconds. The distance a is 12.0 cm, and b is 15.0 cm.

(Figure 1)

Express your answer in terms of the variable t. Express all numbers in your answer using two significant figures.

Seite 6

Problem 29.29

In the figure, the rod moves to the right with a speed of 1.0 m/s and has a resistance of 3.0 Ω .(Figure 1) The rail separation is l = 22.0 cm . The magnetic field is 0.39 T , and the resistance of the U-shaped conductor is 26.5 Ω at a given instant.

Calculate the induced emf.

Express your answer to two significant figures and include the appropriate units.

The induced emf is calculated as follows

e=Blv

e=emf

B=magnetic field

l=separation

v= speed

=(0.39 T)(0.22 m)(1.0 m/s)

=0.0858V

Calculate the current in the U-shaped conductor.

Express your answer to two significant figures and include the appropriate units.

The current flowing in the U-shaped conductor is

I= e/r+R

I= current

e=emf

r=resistance of the rod

R=resistance of the U-shaped conductor

= (0.0858V)/(3.0 ohms) + (26.5 ohms))

=2.91*10^-3 A

=2.91 mA

Calculate the external force needed to keep the rod's velocity constant at that instant.

Express your answer to two significant figures and include the appropriate units.

The external force on the rod is

F=BIl

F= external force

B= magnetic field

I= current

l= separation

=(0.39 T)(2.91*10^-3)(0.22 m)

=2.5*10^-4 N

Seite 7

Problem 29.40

A 320-loop circular armature coil with a diameter of 12.8 cm rotates at 100 rev/s in a uniform magnetic field of strength 0.46 T .

What is the rms voltage output of the generator?

Express your answer using two significant figures.

Peak voltage= wNBA

angular frequency (w)= 100rev/s

100*2pi rad/s

628.32 rad/s

no of turns in the armature (N)= 320

Area of one loop (A)= pi r^2

3.13*((12.8*10^-2)/2)^2

1.4*10^-2 m^2

Magnetic Field (B)= 0.46 T

V0= (628.32)(320)(1.4*10^-2)(0.46)

=somehow it got 840 V

What would you do to the rotation frequency in order to double the rms voltage output?

We would double the rotation frequency to 200 rev/s

Seite 8

Problem 29.54

In a circular region, there is a uniform magnetic field B⃗ pointing into the page (see the figure). An xy coordinate system has its origin at the circular region's center. A free positive point charge +Q = 1.0 μC is initially at rest at a position x = + 10 cmon the x axis.

(Figure 1)

If the magnitude of the magnetic field is now decreased at a rate of -0.29 T/s , what force (magnitude and direction) will act on +Q?

Express your answer using two significant figures.

The electric field due to change in this magnetic field is

E=(x/2)dB/dt

x=+10m

dB/dt= change in magnetic field= -0.29 T/s​​​​​​​

=-.0145N/C

Thus the force is

​​​​​​​​​​​​​​F=qE

​​​​​​​F= force

q=charge

E=electric field

=(1.0*10^-6C)(-.0145 N/C)

-1.45*10^-8N

But positive

15nN

The flux is pointing in and decreasing. By Lenz's law, the induced E field is going to try to make charges move so as to oppose that change. Your positive charge that is on the x-axis is going to want to move clockwise, in the negative y direction; that, by the right hand rule, will make its B field towards the centre of the loop point in to the page, counteracting the other change. Thus, the x component of force is 0 and the y component of force is 9.0 n and is directed in the

NEGATIVE Y AXIS

Seite 9

Problem 29.63

A pair of power transmission lines each have a 0.75 Ω resistance and carry 760 A over 9.0 km. If the rms input voltage is 30 kV .

Calculate the voltage at the other end.

Express your answer using two significant figures.

Double the resistance of a single line to calculate total resistance-

R=2R

R=resistance

=2(0.75ohms)

=1.5ohms

Use Ohm's law to calculate voltage drop

Vdrop= IR

​​​​​​​I=current

R=resistance

=(760A)(1.5ohms )

= 1140V

= 1.14kV

Write the expression of output voltage

​​​​​​​V(output)=V(input)-V(drop)

=30kV- 1.14kV

​​​​​​​= 28.86 kV

Calculate the power input.

Express your answer using two significant figures.

Write the expression for input power

P(input)=V(input)I

​​​​​​​I=current

=(30kV)(760A)

​​​​​2.3*10^7 W

Calculate power loss in the lines.

Express your answer using two significant figures.

Write the expression for power loss

P(loss)=V(drop)I

=(1.14kV)(760A)

​​​​​​​=8.7*10^5 W

Calculate the power output.

Express your answer using two significant figures.

P(output)=Voutput I

Seite 10

Problem 29.67

A coil with 140 turns, a radius of 5.0 cm, and a resistance of 12Ω surrounds a solenoid with 230 turns/cm and a radius of 4.3 cm ; see the figure. The current in the solenoid changes at a constant rate from 0 to 1.7 A in 0.12 s.

Calculate the magnitude and direction of the induced current in the 140-turn coil.

Express your answer using two significant figures.

A magnetic field inside a solenoid

B=u(0)ni

​​​​​​​B= magnetic field

u(0)=permeability

n= numer of turns

i= current

Flux through the coil

​​​​​​​flux=NBA​​​​​​​

​​​​​​​N= number of turns

A= area of cross section of solenoid=pir^2

Induced emf= rate of change in flux

​​​​​​​​​​​​​​e=(d/dt)(NBA)

e=(d/dt)(u(0)niA)

​​​​​​​e=(di/dt)(Nu(0)nA)

N=

n=

(di/dt)=

A=

R=

Seite 11

Transformers

Seite 12

A Driven Series L-C Circuit

Seite 13

A Parallel and Series LR Circuit CQ

Seite 14

Alternating Voltages and Currents

Seite 15

Problem 30.1

A 2.47 m -long coil containing 225 loops is wound on an iron core (average μ=1850μ0) along with a second coil of 115 loops. The loops of each coil have a radius of 2.40 cm . If the current in the first coil drops uniformly from 11.0 A to zero in 94.0 ms.

Determine the mutual inductance M.

Determine the emf induced in the second coil.

Seite 16

Problem 30.11

To demonstrate the large size of the henry unit, a physics professor wants to wind an air-filled solenoid with self-inductance of 1.0 H on the outside of a 15 cm diameter plastic hollow tube using copper wire with a 0.83 mm diameter. The solenoid is to be tightly wound with each turn touching its neighbor (the wire has a thin insulating layer on its surface so the neighboring turns are not in electrical contact).

How long will the plastic tube need to be?

Express your answer to two significant figures and include the appropriate units.

self-inductance of an air-filled solenoid

L=[(permeability)(N^2)(A)]/l

​​​​​​​N=number of turns

A=cross-sectional area of wire

l=length of wire

​​​​​​​

How many kilometers of copper wire will be required?

Express your answer to two significant figures and include the appropriate units.

What will be the resistance of this solenoid?

Express your answer to two significant figures and include the appropriate units.

Seite 17

Problem 30.16

Typical large values for electric and magnetic fields attained in laboratories are about 1.0×104V/mand 2.0T.

Determine the energy density for electric field.

Express your answer to two significant figures and include the appropriate units.

Energy in Electric Field= 0.5epsilonE^2

​​​​​​​E= 1*10^4

(.5)(8.85*10^-12)(1*10^4)

4.425*10^-4 J/m^3​​​​​​​

Determine the energy density for magnetic field.

Express your answer to two significant figures and include the appropriate units.

Energy in magnetic field= .5(B^2/uthingy)

B=2.0T

.5(4/pi*10^-67

​​​​​​​=1591549.43 J/m^3

What magnitude electric field would be needed to produce the same energy density as the 2.0-T magnetic field?

Express your answer to two significant figures and include the appropriate units.

WE=WB

.5epsilonE^2=.5(B^2/uthingy)

E^2=B^2/uthing*epsilon

E=sqr[B^2/uthing*epsilon

E=2/sqr[4pi*10^-7*8.85*10^-12]

E=0.6*10^9 v/m

Seite 18

Problem 30.26

In the circuit on the figure, determine the current in each resistor (I1,I2,I3) at the following moments:

When the switch is closed.

Express your answers in terms of some or all of the variables E, R1, R2, R3. Enter your answers separated by commas.

E=emf

I(3)=0

I(1)=I(2)

I(1)r(1)+I(2)r(2-E=0

since I(1)=I(2)-

I(1)[r(1)+r(2)]=E

I(1)=E/[r(1)+r(2)]

I(1)=I(2)=E/[r(1)+r(2)], I(3)=0

A long time after the switch is closed.

Express your answers in terms of some or all of the variables E, R1, R2, R3. Enter your answers separated by commas.

I(1)=E/r(eq)=[E[r(2)+r(3)]]/[r(1)r(2)+r(1)r(3)+r(2)r(3)]

Er(3)/[r(1)r(2)+r(1)r(3)+r(2)r(3)]

Er(2)/[r(1)r(2)+r(1)r(3)+r(2)r(3)]

After the switch has been closed for a long time, and then reopened, what is each current just after it is opened?

Express your answers in terms of some or all of the variables E, R1, R2, R3. Enter your answers separated by commas.

After the switch has been closed for a long time, and then reopened, what is each current after a long time?

Express your answers in terms of some or all of the variables E, R1, R2, R3. Enter your answers separated by commas.

Seite 19

Problem 30.45

What is the reactance of a 8.6×10−2 μF capacitor connected to a 28 kV (rms), 680 Hz line?

Express your answer using two significant figures.

Capacitive reatance

​​​​​​​X(c)=1/2pifC

=1/2pi(8.6*10^-2*10^-6)(640))

=2891.6ohm

Determine the frequency and the peak value of the current.

Express your answer using two significant figures.

Frequency=680Hz

Peak value of current

I(max)=Vmax/Xc

V(max)=1.414*V(rms)

=1.414*2800

=39592V

=39592/2891.6

=13.69A

Seite 20

Problem 30.52

What is the rms current in a series RC circuit if R = 4.1 kΩ , C = 0.75 μF , and the rms applied voltage is 140 V at 45 Hz ?

Express your answer using two significant figures.

The rms current is-

​​​​​​​I(rms)= V(rms)/Z

​​​​​​​Z= the impedance of RC circuit

I(rms)=V(rms)/sqr(R^2+X(c)^2)​​​​​​​

X(c)= capacitive reactance

I(rms)=V(rms)/sqr(R^2+(1/(wC))^2)

​​​​​​​w=angular frequency

I(rms)=V(rms)/sqr(R^2+(1/(2pifC))^2)

​​​​​​​f = frequency

Plug it in

​​​​​​​​​​​​​​140V/sqr((4100^2+(1/(2pi(45Hz)(.75*10^-6F))^2

I(rms)=0.022A

What is the phase angle between voltage and current?

Express your answer using two significant figures.

phase angle formula

​​​​​​​=tan^-1(X(c)/R)

=tan^-1((1/(2pifC)/R)

Plug it in

​​​​​​​​​​​​​​tan^-1((1/2pi45(.75*10^-6))/4100)

49degrees BUT

check sign

-49*

What is the power dissipated by the circuit?

Express your answer using two significant figures.

P=I(rms)^2 R

2.1W

What are the voltmeter readings across R and C?

Express your answer using two significant figures.

To calculate the voltmeter reading across a resistor-

V(R)=I(rms)R

=(0.022)(4100)

=90.2

To calculate the voltmeter reading across a capacitor

​​​​​​​​​​​​​​V(R)=I(rms)X(c)

=I(rms)(1/(2pifC))

=(0.022)(1/2pi45.75*10^-6))

=103.7

Seite 21

Problem 30.60

An LRC series circuit with R = 140 Ω , L = 11 mH , and C = 1.3 μF is powered by an ac voltage source of peak voltage V0 = 530 V and frequency f = 480 Hz .

Determine the peak current that flows in this circuit.

Express your answer using two significant figures.

The rms current is-

​​​​​​​I(rms)= V(rms)/Z

​​​​​​​Z= the impedance of RC circuit

I(rms)=V(rms)/sqr(R^2+X(c)^2)​​​​​​​

X(c)= capacitive reactance

I(rms)=V(rms)/sqr(R^2+(1/(wC))^2)

​​​​​​​w=angular frequency

I(rms)=V(rms)/sqr(R^2+(1/(2pifC))^2)

​​​​​​​f = frequency

Plug it in

​​​​​​​​​​​​​​540V/sqr((140^2+(1/(2pi(480Hz)(1.8*10^-6F))^2

I(rms)=A

Determine the phase angle of the source voltage relative to the current.

Express your answer using two significant figures.

Determine the peak voltage across R.

Express your answer using two significant figures.

Determine the phase angle of the voltage across R relative to the source voltage.

Express your answer using two significant figures.

Determine the peak voltage across L.

Express your answer using two significant figures.

Determine the phase angle of the voltage across L relative to the source voltage.

Express your answer using two significant figures.

Determine the peak voltage across C.

Express your answer using two significant figures.

Determine the phase angle of the voltage across C relative to the source voltage.

Express your answer using two significant figures.

Seite 22

Problem 30.72

At t = 0, the current through a 58.5 mH inductor is 58.5 mA and is increasing at the rate of 68.0 mA/s .

What is the initial energy stored in the inductor, and how long does it take for the energy to increase by a factor of 3.5 from the initial value?

Express your answer using three significant figures.

Formula for energy stored in an inductor

​​​​​​​E=(1/2)LI^2

​​​​​​​L=inductnce

I=current

E=energy stored in an inductor

=(1/2)(58.5*10^-3)(58.5*10^-3)^2

=1.001*10^-4 J

Calculate the energy increased by factor of 3.5 from the initial value-

​​​​​​​E1=(1.001*10^-4J)3.5

=3.5035*10^-4

The new current in the inductor is

​​​​​​​E1=(1/2LI^2

I=sqr((2E/L))

=sqr(2(3.5035*10^-4J)/58.5*10^-3

=0.10944mA

The difference in current is

​​​​​​​​​​​​​​DI=0.10944mA-58.5mA

The current increases by 68.0 mA/s and the time within that the current by 50.94mA is-

​​​​​​​t=50.94mA/68mA/s

​​​​​​​0.749s