suggested+solution+fybch20 21nvc06+magnetic+induction+&+ac+circuits

8/3/2019 Suggested+Solution+FyBCh20 21NVC06+Magnetic+Induction+&+AC+Circuits

1/12

Suggested solution FyBCh20-21, 24 NVC06 magnetic Induction & AC circuits NV-College

[email protected] to use for educational purposes. Not for sale. 1/12

Part I: Answer part I on this paperPart I: Answer part I on this paper

1. In the figure below a proton is projected horizontally at sm / into the regionbetween two parallel charged plates which are

100.55

mm0.10 apart. The potential difference

between the plates is V.600 .

a) Draw the electric filed lines as clear as possible on the figure. [1/0]Data: mmmd 2100.10.10 == , VV .600= , ,smv x /100.5

50 =

Suggested answer:

The electric field is uniform in the region

between the plate. Its direction is from

positive to negative, as illustrated in the

figure.

b) In which direction will the proton be deflected? [1/0]i)

Toward the bottom of the paper.ii) Toward the top of the page.

iii) Into the page.iv) Out of the page.

Answer: Alternative: i.

The proton will be deflected towards the bottom of the paper. This may be explain either by

using the fact that the positive charges, and therefore protons, are attracted to the negative

charges. We may also use EeEQFrrr

== to explain the effect, i.e. in the region between the

plates; there is a force on the incoming protons in the direction of the electric field.

c) What is the magnitude of the electric field? [1/0]i) mV /0.60 ii) mmV /0.60 iii) mV /0.60 iv) mkV /0.60 v) mMV /0.60

Suggested solution: Answer: Alternative v: mkVE /0.60=

Data: mmmd 2100.10.10 == , V.600V = ;d

VEdE ==V

mkVmVmVm

V

E /0.60/1000.6/10600100.1

.600 422

===

= ++

++++++++++++

++++++++++++

pmm0.10

cm0.2

sm /100.5 5


2/12



d) The magnitude of the electric force on the proton is: [1/0]d) The magnitude of the electric force on the proton is: [1/0]i)i) N.106.9 14 ii) N.106.1 19 iii) N15106.9 iv) N15106.2

Suggested solution: Answer: Alternative iii: NF 15106.9 =

Data: mmmd 2100.10.10 == , V.600V = ; mkVE /0.60=

EeEQFrrr

== NmVF 15319 106.9/100.60106.1 + ==

e) If the plates are cm0.2 long, and the proton is ejected horizontally into the plates,exactly in the middle of the plates, i.e. mm0.5 from each plate, in what angle

will the proton leave the plates? [1/0]

i) at = 25 with x-axis. Downward.ii)

at

=13 with x-axis. Downward.iii)at = 3.1 with x-axis. Downward.

iv)at = 25 with x-axis. Upward.v) It will not leave the plates. It will hit the plate below.vi)at = with x-axis. (fill the space, if none of above corresponds to your

calculations. Indicate if it is DOWNWARDS or UPWARDS.)

Draw the required figures, and show the details of your calculations: [0/3]

Suggested solution: Answer: Alternative i: at = 25 with x-axis.

Downward

Data: mmmd 2100.10.10 == , VV .600= ; mkVE /0.60= ,

smvx /100.5

5

0 = , ,cmx 0.2= mmmy3

max 100.50.5== , NF 15106.9 =

This is a projectile problem. The proton has a constant velocity in thehorizontal direction: skmv x /.1000 = , while in the vertical direction, its

initial velocity is 00 =yv , while its is facing an downward force of

NF 15106.9 = and therefore, accelerating at the rate of

amFrr

= 21227

15

/1074.51067.1

106.9sm

kg

N

m

Fa

p

y

y =

==

During the time that the proton moves horizontally , i.e.

during

mx2

100.2 =

ssm

m

v

xt

8

5

2

100.4/100.5

100.2 === it moves downward :m31060.4

( )( ) mtay y 328122

1059.4100.41074.52

1

2

1 ===

Due to the fact that , the proton will leave the

field region.

mmy33 1000.51059.4


3/12



To find the angle at which the proton exits the field region, we need tofind the velocity in the vertical direction at the time the proton is leavingthe region:

To find the angle at which the proton exits the field region, we need tofind the velocity in the vertical direction at the time the proton is leavingthe region:

( )( ) smtav yy /100.41074.5 812 ==

smsmvy /103.2/10296.255 =

at = 25 as illustrated below:

=

=

sm

sm

v

v

x

y

/100.5

/10296.2tantan

5

511

( ) == 2566.244592.0tan 1 Note that this angle is very different from that of:

=

=

=

9.12

100.2

1059.4tantan

2

311

m

m

x

y

Answer: Alternative i: at = 25 with x-axis. Downward

f) If we would like the ejected proton leaving the plates area horizontally, we mayturn on a magnetic field in the region. The direction of such a magnetic field is:

[1/0]

i) perpendicular to the original direction of the proton, into the paper.ii) perpendicular to the original direction of the proton, out of the paper.iii)perpendicular to the original direction of the proton, vertically upward.iv)perpendicular to the original direction of the proton, vertically downwardv)

horizontal to the left in the original direction of the protons motion.vi)horizontal to the right in the opposite direction of the protons motion.

Answer: Alternative __________________

Draw necessary figures and explain your reasoning. [0/2]

Suggested solution: Answer: Alternative ii: out of the paper

The magnetic force must be identical to the electric force on the moving proton, but in the

opposite direction. Therefore, the direction of the magnetic force must be upwards. Due to the

fact that the charge of proton is positive and it is originally moving horizontally to the left, the

direction of the uniform magnetic field in the region between the plates must be out of the

paper as illustrated in the figure below:

++++++++++++

B

NFyB 15106.9 =

NFyE15106.9 =

smvx

/100.5 50 =

TB 12.0=

proton

= 9.12

25


4/12



g) What is the magnitude of such a magnetic field: [1/0]g) What is the magnitude of such a magnetic field: [1/0]i)i) TB 12= ii) mTB 12= iii) TB 12.0= iv) kTB 12= v) MTB 12= vi)None of above. It is :__________

Answer: Alternative __________________

Show the details of your calculations: [0/2]

Answer: Alternative iii: TB 12.0=

Data: mmmd 2100.10.10 == , VV .600= , ,smv x /100.55

0 =

As mentioned above, the magnitude of the magnetic force exerted by the magnetic field mustbe exactly equal to the magnitude of the electric force NFy

15106.9 = but in the opposite

direction. Therefore, using BvQFrrr

= , and EB FFrr

= , andd

VEdEV ==

EevBe /=/ EvB = v

EB =

dv

V

v

EB

==

dv

VB

=

Tdv

VB 12.0

100.1100.5

60025=

=

=


5/12



decreasesI

2. Show clearly the direction of the induced current in the circular loops below due to thecurrent shown? Why? Explain and Draw it on the figure:

i. Explanation: [1/1]The direction of the induced current in the r

is counter clock-wise. This is due to the factthat the direction of the magnetic field due to

the wire inside the ring is out of the paper, but

because the current in the wire decreases, its

magnetic field also decreases, and therefore,

the magnetic flux through the ring also

decreases. Therefore, there will be an induced

emf in the circular loop (due to the Lenss law)

such that it opposes the decrease in the magnetic flux, and therefore,

the direction of the induced current in the loop will be counter clock wise because this

type of current will produce a magnetic field inside the loop which is out of the paper.

ing

nduced

f

e

ii. Explanation: [1/1]The direction of the i

current in the ring is counter

clock-wise. This is due to

the fact that the direction o

the magnetic field due to th

wire inside the ring is into

the paper, but because the current in the wire increases,

its magnetic field also increase, and therefore, the magnetic flux through the ring also

increase. Therefore, there will be an induced emf in the circular loop (due to the Lenss

law) such that it opposes the increase in the magnetic flux. Therefore, the direction of

the induced current in the loop will be counter clock wise because this type of current

will produce a magnetic field inside the loop which is out of the paper.

3. The magnetic flux through a coil of wire containing 100 loops change from Wb56 toWb54 in s20.0 . What is the emf induced in the coil? [2/0]

Suggested solution:

Data: 100=N , WbB 561 = , WbB 542 = , st 20.0= , Problem: ?=

Answer: The emf induced in the coil is kV55= .

dt

dN B

=

tN

BB

= 12

( )kVV 55105.5

2.0

5654100 4 ==

=

The positive sign demonstrates the fact that the induced emf opposes the change

(decrease) in the magnetic flux (Lenses law.)

increasesI

decreasesI

increasesI


6/12



4. An electron and an alpha particle have the same kinetic energy upon entering a region ofconstant magnetic field. What is the ratio of the radii of their circular paths? The

4. An electron and an alpha particle have the same kinetic energy upon entering a region ofconstant magnetic field. What is the ratio of the radii of their circular paths? The

particle is the nucleus of He atom and has a charge of e+ 2 and mass of

kgkgmHe-27-27 1065.6106605.1002602.4u002602.4 == [1/2]

Suggested solution: Answer:He

e

He

e

m

m

r

r2= ; 43

e

He

r

r

Data:Pe KEKE

EE = , Ce 19106.1 = , kgkgme-31-31 109.11109.10938188 = ,

kgkgmHe-27-27 1065.6106605.1002602.4u002602.4 ==

Problem: ?=P

e

r

r

Answer:

BvQFrrr

=

==

=

r

vmmaF

BvQF

CC

2 r

vmBvQ

2/

=/

BQ

vmr

=

Assuming that both the electron and Alpha

particle were originally moving at the same

initial kinetic energy in an inertial frame,

i.e.: 22

2

1

2

1HeHeee vmvm

/=

/

e

He

He

e

mm

vv =2

2

e

He

He

e

mm

vv =

BQ

vmr

=

He

e

He

e

HeHe

ee

HeHe

ee

He

e

v

v

m

m

vm

vm

Be

vm

Be

vm

r

r=

=

//

//

=22

2

He

e

e

He

He

e

He

e

m

m

m

m

m

m

r

r22 ==

Answer:He

e

He

e

m

m

r

r2=

437.426.72952

1

109.10938188

106463206.6

2

131-

-27

==

=

kg

kg

r

r

e

He Answer: 43e

He

r

r

Note that due to their opposite charges, Alpha particles and electrons path are in two

opposite directions as illustrated above.

B

eev

pv

pv e+


7/12



5. In a circuit a resistance of resistor .500 is connected in series to a capacitor ofcapacitance F0.25 and an inductance mH250 and a sinusoidal alternating emf

device operating at Hz50 and amplitude V0.25 .

a)

What is the current amplitude in the circuit? [2/2]b) What is the potential difference as a function of time across the inductance? [1/1]c) What is the phase constant of the current in the circuit relative to the driving

emf? [1/1]

d) What is the current in the circuit as a function of time? [1/1]Suggested solution:

Answer: (a) mAI 2.50max = , (b) ( ) VttVL

2100sin95.3

; (c) 57.5 ,

(d) ( ) ( ) mAttI 21077.8100sin2.50 += ; VV 25max =

Data: = .500R , FFC6

100.250.25

== , HmHL3

10250250

== ,Hzf 50= , V0.25max =

Problem:

Hzf 50= === 1005022 f = 100

The reactance of the capacitoris

==6100.25100

11

CXC =

2104

CX

The reactance of the inductance is == 310250100 LXL = 25LX

The circuits impedance is ( ) =

+=+= 2

2

2221002.5

40025500

CL XXRZ

=21002.5Z

mAAV

ZI 98.41098.4

1002.5

0.25 22

maxmax ==

==

mAI 98.4max =

VVILIXV LL 91.31098.4252

maxmaxmax === VVL 91.3max

VVIC

IXV CC 34.61098.41041 22

maxmaxmax =

==

VVC 34.6max

VVIRVR 9.241098.45002

maxmax == VVR 9.24max

The current in passing the inductance lags the voltage by rad2

90

= . Therefore

( ) ( ) VtILtIXtIXtV LLL

=

==

2100sin

2100sin maxmax

( ) VtVttVL

=

2100sin95.3

2100sin1002.525 2

( ) VttVL

2100sin95.3


8/12


The phase angle isThe phase angle is

radR

XX CL 2111 1077.857.5500

8.48tan

500

40025

tantan =

=

=

=

The negative phase constant 57.5 is an indication that the load is mainly capacitive,

andLC XX > . Therefore, the current leads the driving emf by 57.5 or by rad

21077.8 .

Input ( ) ( ) ( ) ( ) VttttV === 100sin25sinmax

Output ( ) ( ) ( ) mAttItI 2max 1077.8100sin98.4sin +==

( ) ( ) mAttI 21077.8100sin98.4 +=

VVR 9.24max

VVC 34.6max

VVL 91.3max

VV 25max =

t

57.5



9/12



CD

AB

E

F

18 cm

In the figure below the rod moves at a speed of sm /5.2 , is cm0.18 long, and has a

resistance 5.0 . The magnetic field is T80.0 and

the resistance of the U-shaped conductor is

5.22 at a given instant. Calculate:

a.

the induced emf, [2/1]b. the current flowing in the circuit, and[2/3]

c. the external force necessary to ensure that the rod is moving atconstant velocity

at the instance. [2/3]

Suggested solution:

Data: smv /5.2= , cm0.18=l , = 5.0rodR , TB 80.0= , = 5.22R

Problem: ?= , ?=I , ?=F

The emf induced in the circuit is V36.0 from C to B.

The current AmI 4.14= from C to B is induced in the closed circuit.

An external force of NF 31007.2 = is needed to ensure that the rod is moving at a

constant velocity smv /5.1= .

( ) ( ) ( )vB

dt

dxB

dt

xdB

dt

dAB

dt

BAd

dt

ABd

dt

d Bll

lrr

==

===

=

=

VvB 36.05.218.080.0 === l V36.0 (2/1)

Due to the fact that the magnetic flux decreases (because the area is decreasing) the

induce emf will oppose the reduction in the magnetic flux. Therefore, the current is from

the point C to B (downward) which produces a magnetic field inside the loop in the

same direction as the original one (i.e. into the page.)

( )IRR rodU += ( )rodU RRI

+=

( )AmAI 4.141044.1

5.25.22

36.0 2 ==+

=

AmAI 4.141044.1 2 ==

[2/3]

BIFr

lrr= BIBIF ll == sin NF 32 1007.280.018.01044.1 ==

mNNF 07.21007.2 3 == [2/3]

6. Two first-order spectrum lines are measured by an 8500-line/cm spectroscope at angles,on each side of centre, of 8326 + , 8041 + and 8426 , 9141 . What are thewavelengths? (1/4)

Suggested solution:

Data: = md sin ; mcm

d6

101.188500

1 == ; First order 1=m :

=+=

+=+

+=++

= 26.71734262

6826

2

848326

2

842683261 ; ?1 = (0.5/0.5)

=+=

+=++

= 41.2255.13412

7241

2

914180412 ; ?2 = (0.5/0.5)

= md sin


10/12



( ) nmmd 529100.0052926.717sin85

10sin 4

4

11 ====

(0/1.5) Green nm5291 =

( ) nmmd 775100.0077541.225sin85

10sin 4

4

22 ====

(0/1.5) Red nm7752 =


11/12



7. Two stiff parallel wires a distance7. Two stiff parallel wires a distance l apart in a horizontal plane act as rails to support alight metal rod of mass m that is parallel to the surface of the ground and perpendicular

to the parallel rails as illustrated

below.

A uniform magnetic field B

directed vertically downward (intothe page in the figure) acts

throughout. The system is connected

to a electric power source that

provides the system with a constant

current I .

a) In which direction does therod move? Why? Explain!

[1/1]

i) East.ii) Westiii)North.iv)Sought.v) Up.vi)Down.

Suggested solution: Answer: Alternative: ii: West; tm

BIv =

l; t

m

mgBIatv k

==

l

The magnetic force on a current I in a magnetic field B is BIFr

lrr= . Using the right

hand rule we may recognize that there is a force to the left, i.e. to the west on the mobile

rod, and therefore, the rod moves to the left. It accelerates to the left at the rate of

mBIa l= .

Calculate the speed of the rod as a function of time [2/4/]

b) if the rails are considered frictionless.c) if the coefficient of the friction between the rod and the trails is

k .

Suggested solution: Answer: tm

BIv =

l; t

m

mgBIatv k

==

l

As discussed above, the rod accelerates at the rate ofm

BIa

l= on a frictionless surface,

and if the initial velocity of the rod is assumed zero, its velocity as a function of time

may be expressed as:

+= atvv 0 tm

BIv =

l

If the coefficient of the kinetic friction between the rod and the rail isk , the friction

force may be expressed as mgFF kNkf == and it is in the opposite direction of the

motion and the resultant force on the rod may be expressed as:

=

==

maF

mgBIFFF kfB l

m

mgBIa k

=

l t

m

mgBIatv k

==

l

NORTH

WEST EAST

SOUGHT

NORTH

WEST EAST

SOUGHT

BIFr

lrr= B

I


12/12



8. A mass spectrometer is being used to monitor air pollutants. It is difficult, however, toseparate molecules with nearly equal mass such as CO ( u0106.28 ) and 2N

( u0134.28 ). How large a radius of curvature must a spectrometer have if these two

molecules are to be separated on film by mm00.1 . (2/4/)

Suggested solution:Data: umCO 0106.28= , umN 0134.282 =

mmmrrd NCO31000.100.122

2

=== ,

Problem: ?=r

Answer: The radius of curvature of the mass spectrometer must be m00.5r

In the mass spectroscope ions are produced by heating, or by an electric current. These

ions are directed through a relatively small hole to a region known as velocity selector

where there are both electric field Er

(pointing up) and magnetic field1Br

(pointing out

of the paper toward the reader.) The magnitudes of the fields are adjusted such that those

ions whose velocity fits, 1vBQEQ /=/ , i.e. 1B

E

v = will follow a straight path passing

through a second hole and entering a new region where magnetic field2B

r(pointing out

of the paper toward the reader) exists. The ions will then follow a circular path of radius

r satisfyingr

vmQvB

2

2 = . The radius r may be measured using a photographic plate

where ions strike and darken the film:

=

=

r

vmQvB

QvBQE

2

2

1

=

=

r

v

mQB

B

Ev

2

1 1

12

rB

Em

r

B

E

mr

vmQB ===

21BQB

Emr=

Due to the fact that ions are assumed having the same chare (for example ionized once)

sharing identical electric and magnetic fields Er

, 1Br

and 2Br

we may express the radius

of curvature of each beam as

=

=

21

2122

BQB

Emr

BQB

Emr

COCO

NN

Dividing the equations to each other may result in:CO

N

CO

N

m

m

r

r22 =

CO

CO

N

N rm

mr 2

2= ( ) CO

CO

N

CON rm

mrrd

== 122 2

2

=

12 2

CO

N

CO

m

m

dr

m

u

u

mrr CO 00189.5

10106.28

0134.282

1000.1 3=

=

mmrr CO 00.500189.5 =

suggested+solution+fybch20 21nvc06+magnetic+induction+&+ac+circuits

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