Suggested+Solution+VGMVG LevelFyBCh20 21NVC08+Magnetism,+Magnetic+Induction,+AC

8/3/2019 Suggested+Solution+VGMVG LevelFyBCh20 21NVC08+Magnetism,+Magnetic+Induction,+AC

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Solution VG/MVG-LevelFyBCh20-21NVC08 Magnetism & Magnetic Induction NV-College

© [email protected] ☺Free to use for educational purposes. Not for sale. 1/8

FyBCh20-21NVC08

VG/MVG-Level Test On Magnetism and Magnetic InductionInstructions:

The Test Warning! There are more than one version of the test.

Tools Approved formula sheets, ruler, and graphic calculator. You may use your personalized

formula booklet which has your name on it. This should be submitted along with the

test.

Time: 10:40-12:05;

The multi-choice problems must be answered on the this paper. The solutions to each

question must also be answered in the space provided. Only, if you need additional

space you may write your solutions in an additional page that will be provided if

needed.

Limits: There are 6 problems in this test and it gives at maximum 27 points, 17 of which areVG points. Problems number 4, 5 and 6 are of the greatest important for MVG.

Pass (G): minimum 9 points

VG: G+ in the G test. V/MVG test: Minimum of total 18 points , 6 of which are VG points

MVG: G+ in the G test. V/MVG test: Minimum of total 21 points , 12 of which are VG points

and MVG ¤ quality in all problems specially on the problems 4, 5 and 6.

Please answer as clear as possible. Do not forget the UNITS!

Enjoy it! Behzad

B LRC emf Gener. LR Ind. t I ind Induct Summin

G min

VG min

MVG

1 2a 2b 3 4 5a 5b 6a 6b 27 9 18 21

G 1 1 1 1 0 1 2 2 1 10

VG 2 1 1 2 3 3 1 1 3 17 6 12

MVG 1235¤ M25¤ 1235¤ M1235¤

G

VG

MVG

MVG- quality ¤ Prob. 4 Prob. 5 Prob. 6 Other

Problems

M1 Formulates and develops the problem, uses

general methods with problem solving.

M2 Analyses and interprets the results, concludesand evaluates if they are reasonable.

M3 Carries out mathematical proof, or analysesmathematical reasoning.

M4 Evaluates and compares different methodsand mathematical models.

M5 The presentation is structured, and

mathematical language is correct.





1. In the accompanied figure the

graph of a uniform magnetic

field, t B through a conducting

loop is plotted as a function of

time. The direction of themagnetic field is perpendicular to

the plane of the loop. Rank the

six regions of the graph

according to the magnitude of

emf induced in the loop, the

smallest first. Why? Explain.[1/2]

0

20

40

60

80

100

120

140

0 3 6 9 12 15 18 21 24 27 30

t [s]

B [ m T ]

Region a b c d e f

Time st 60 st 96 st 129 st 1512 st 2115 st 2721

Suggested solution:Ignoring the sign of the change in the magnetic flux, we may construct the

following table. Note that the induced emf in both regions b and d are zero.

Similarly the magnitude of the change in the magnetic flux in both regions e

and d are identical.

0

t

B 0

t

B 10

t

B 10

t

B 20

t

B 40

t

B

b d e d a c

st 96 st 1512 st 2115 st 2721 st 60 st 129

2. A resistor, a capacitor, and an inductor are connected in series to an AC power sourceof frequency f . The effective voltages across the circuit components are V V R 0.12 ,

V V C 0.18 , and V V L 0.14 .

a. Find the effective voltage of the source. [1/1]

b. Find the phase angle in the circuit. [1/1]

Suggested solutions:

Data: V V R 0.12 , V V C 0.18 , and V V L 0.14 .

a. Answer: The effective voltage of the source is V V rms 7.11 . [1/0]

2

00

2

00 C L R V V V V

22

Crms Lrms Rrmsrms V V V V

V V V rms 6.121604120.140.18122222 [0/1]

b. Answer: The phase angle in the circuit is 31 . Voltage lags the current

by 31 . [1/0]

R

X X

V

V V C L

R

C L

0

00tan

4.183

1tan3

1

0.12

0.4

0.12

0.180.14tan1

0

00 R

C L

V V V [0/1]





3. The armature of a Hz50 ac generator rotates in a T 25.0 magnetic field. If the area

of the coil is 2400 cm . How many loops must the coil contain if the peak output is to

be V .3000

? [1/2]

Suggested solution:The emf is given by:

turns

BA N

A B N t A B N

9549.955021040025.0

.300

sin

4

0

max0

[1/2]

4. The potential difference across a given coil is V 5.22 at an instant when current is

mA860 and is increasing at a rate s A / 40.3 . At a later instant, the potential

difference is V 2.16 whereas the current is mA700 and is decreasing at a rate of

s A / 80.1 . Determine the inductance and resistance of the coil. [0/3/ M1235¤ ]

Suggested solution:The voltmeter across the coil registers a potential which is equivalent to asystem of “pure” inductor with inductance L connected in series with aresistor of resistance R :

dt

dI L RI V

Therefore, the following equation systems must be satisfiedsimultaneously:

2.1680.1700.0

5.2240.3860.0

L R L R

which may be solved by eliminating first L by multiplying the firstequation by 80.1 and multiply the second equation by 40.3 and then

subtracting the resultant equations as:

40.32.1640.380.140.3700.0

80.15.2280.140.380.1860.0

L R

L R

3.24333.24

928.3

58.9558.95928.3

08.5512.6380.2

50.4012.6548.1 R R R

L R

L R

We may determine the value L by substituting the value of 3.24 R in

one of the equations, for example in the first equation:

H L L L

L L L

463.040.3

574.1574.140.3

926.205.2240.35.2240.3926.205.2240.3333.24860.0

Answer: The inductance of the coil is H L 463.0 . It has an internal

resistance 3.24 R .





5. In the accompanied figure a conducting

circular metallic loop of diameter

cm0.50 , and resistance 00.5 is

illustrated. As shown in the figure, fifty

percent of the figure, i.e. the half-circle

lies in a uniform magnetic field t B

that is directed out of the page of the

paper and towards the reader of these

words. The magnitude of the magnetic

field is a quadratic function of time and

may be expressed as

T t t t B 00.900.300.52

The magnetic field B is in Tesla and

time t in seconds. An ideal battery with

emf V ba t 00.9 is connected to the

loop as illustrated in the figure.

a. Find the direction and magnitude of the induced current in the metallic circular

loop by t B

as a function of time t . [1/3]

b. Calculate the direction and magnitude of the current in the loop at st 0.10 ?

Explain. [2/1/ M25¤ ]

Suggested solution:

Data: mcmd 500.00.50 , mmd

r 250.02

500.0

2 ; 00.5 R ,

T t t t B 00.900.300.52 ; V V ba t ba t 00.9 ; st 00.5 ; ?ind I , ? I

a. Answer: At t I 01875.00625.0 ;clockwise

According to Faradays law of induction the magnitude of the inducedelectro motor force ind in a conducting loop is equal to the rate at which

the magnetic flux is changing as a function of time:

dt

A Bd

dt

d Bind

The minus sign is according to the Lenz’ s law which indicates that aninduced current in a conducting loop has a direction such that the

magnetic field due to the induced current always opposes the change inthe magnetic field that is inducing the current. The magnetic flux A B B

through the loop depends on how much of the loop’s area, A , lies in the

magnetic field region B

, and how the area is oriented in the magneticfield.

The area of the half-circle-loop which lies in the magnetic field is2

2r

A

and due to the fact that the direction of the magnetic filed is normal to

the plane of the loop A B A B B

:

T t t t B 00.900.300.52

B





V r

t t dt

d

dt

A Bd induced

200.900.300.5

2

2

[0/1]

V t t r

induced 500.100.5250.000.30.102

2

2

V t induced 09375.03125.0 [0/1]

The negative sign is due to Lenz’s law. The induced current opposes thechange in the magnetic flux. Due to the fact that magnetic flux increasesas a function of time, the induced current opposes the increase in themagnetic flux. Therefore the induced current is clockwise and itsmagnitude is At t I 01875.00625.0 : [0/1]

V t induced 09375.03125.0

Ohm's Law: R

I R

V I I RV ind

ind

At V t

Rt I induced

ind

01875.00625.000.5

09375.03125.0 [1/0]

Answer: At t I ind 01875.00625.0

b. Answer: The net current in the loop at st 0.10 is A I net 58.1

counter-clockwise. [0/1/¤]

A A I ind 02.201875.00.100625.00.10 clockwise [1/0]

It is important to realize that there are two currents in the loop:

i A constant counter-clockwise current due to the ideal battery

60.35

0.18

R

V I ba t . [1/0]

ii A clockwise induced current which is a function of time. Itsmagnitude at st 0.10 is A I ind 02.20.10 .

Therefore the net current in the loop is A I net 58.102.260.3 counter-

clockwise.





6. As illustrated in the accompanying figure a kg00.3 metallic

rod of length cm0.10 is placed on two long parallel conductor

rails which may be assumed resistance free. The conductors are

connected in series to a 250.0 resistor. As shown in the figure a

uniform magnetic field of magnitude T 50.2 is directed out of the

page. The bar is in a good contact with the conductor rails. The

metallic bar is given an initial velocity sm / 00.9 . The friction

coefficient between the rails and the metallic bar is 045.0 .

a. Find the magnitude and direction of the induced current in

the loop immediately after it is shot in the direction as

illustrated in the figure, i.e. towards the top of the page.[2/1]

b. Find the deceleration of the bar at the point its velocity is

dropped to sm / 00.6 . [1/3/M1235¤]

Suggested solution:

Data: T B 50.2 , mcm 100.00.10 , smv / 00.90

, smv / 00.6 ,

250.0 R ; 045.0

a. Answer: V ind 25.2 [1/0]

If the bar moves m x , the area enclosed by the bars, the resistor, and

the metallic rod is m x A . According to Faradays law of induction the

magnitude of the induced electro motor force ind in a conducting loop is

equal to the rate at which the magnetic flux is changing as a function of time:

v Bdt

dx B

dt

xd B

dt

AdB

dt

A Bd

dt

d Bind

[1/0]

The minus sign is according to Lenz’s law. The area decreases bym x A , therefore, the induced emf in the closed circuit resists the

decrease. The induced current in the circuit is therefore counter-clockwise V v Bind 25.200.9100.050.2 [0/1]

Ohm's Law: A

R

I

R

V I I RV ind

ind 00.9250.0

25.2

[1/0]

b. The retarding force consists of two forces: [0/1] i Friction force: N N gmF F N f 33.13257.182.900.3045.0

ii Magnetic force: v I F B

v R

v Bv

Rv I F ind

B

R

v BF B

22

in the opposite direction of motion, i.e. toward the

bottom of the page:

N R

v BF B 60.3

250.0

00.6100.050.22222

N N F F F B Rretard

93.49257.43257.1





22 / 64.1 / 6419.1

00.3

9257.4smsm

m

F a retard

retard [0/1/M1235¤]





MVG- quality ¤ Prob. 4 Prob. 5 Prob. 6 Other

Problems

M1 Formulates and develops the problem, uses

general methods with problem solving.M2 Analyses and interprets the results, concludes

and evaluates if they are reasonable.

M3 Carries out mathematical proof, or analysesmathematical reasoning.

M4 Evaluates and compares different methodsand mathematical models.

M5 The presentation is structured, andmathematical language is correct.

Documents

Suggested+Solution+VGMVG LevelFyBCh20 21NVC08+Magnetism,+Magnetic+Induction,+AC