Final Revisionفيزياء لغات - Colour 2nd

Final Revision A) Essay questions 1. Two identical loops, of copper and aluminum are in a uniform magnetic field perpendicular to their planes (resistivity of copper is less than that of aluminum).

If they are removed from the field rapidly within the same time interval. a. The induced emf generated in the copper loop is ………… (greater than – less than – equal to) the induced emf generated in the aluminum loop. b. In which loop the induced current will be greater? Why?  Answer a. equal b. Copper , because the resistivity of copper is less than that of aluminum 2. The graph represents the electromotive force (emf)

generated in the dynamo. Use the given data to find the average electromotive force through a 1/4 revolution of its coil.  Answer (e.m.f )av = = 3. What would happen to the readings of the instruments indicated in figure, when

the variable resistance R2 is increased?  Answer Req: increases Imain: decreases Reading of ammeter decreases V1 = I R1 V1: decreases V = VB Ir V: increases 4. In the opposite electric circuit , at closing the switch (K): a. The potential of plate…….. increases gradually b. The potential of plate…….. decreases gradually c. When the capacitor charged the reading of ammeter became………..

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d. When the battery replaced with AC source, the potential difference between the capacitor plate has the same phase of ……….  Answer a. A b. B c. Zero d. The source 5. In the opposite circuit: A variable capacitor and an induction coil of no resistance the are connected together with AC source so, XL and effective current is I. If the capacitance decreased to quarter so , and the effective current intensity increased to double, calculate the ratio between  Answer C 2 = C1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) (

)

6. The opposite figure represents the occurrence of

photoelectric phenomena for a certain metallic surface of work function h , what is the velocity of the emitted electrons from the surface of the metal when a light of frequency 2 falls on it? With explanation.  Answer E= E (

) √

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B) Problems 1. A Current of intensity 4 A passed in thin metal wire A When it was connected in parallel with another wire B of the same cross sectional area and from the same material, the intensity of the current was to be increased in the circuit to 6 A to keep the potential difference between metallic wire A remain constant. Calculate the ratio between the lengths of the two wires. ( )  Answer V1 = V2 I1R1 = I2R2 4 R 1 = 2 R2 2 R1 = R2 ( ) ( )

2. From the circuit given in the figure. Calculate:

a. Equivalent resistance.(10 ) b. Current passes in 12 Ω resistor.( 1A)  Answer a. =10  b.

I= I= V6Ω= 3 6 = 18 V I4,8Ω = V12Ω, 24Ω= 1.5 I12Ω=

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=12V

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3. Four electric resistors 2 Ω ,5 Ω ,6 Ω ,8 Ω are connected to a battery of emf 10 V

and internal resistance 1 Ω, it is found that the electric current intensity that flows through 5 Ω resistor double the value of the electric current intensity that flows through 2 Ω a. Show by drawing the way of connecting these resistors together. b. Calculate the current intensity through the battery. (1A)  Answer a.

b. I = I= 4. The following mathematical equations express an electric circuit: I1+I2 = I3 5 = 5I1+ 2.5 I3 25 = 7.5 I2 + 2.5 I3 Draw a simple circuit expresses these equations, then calculate the unknowns in the previous equations  Answer

From equations (1), (2) and (3) using calculator I1=

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, I2=

, I3=

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5. Calculate:

a. Current intensities in each branch. (1.226, -0.516, 0.71) b. Calculate potential difference between (a) and (b) (3.55V)  Answer a. I1+I2 – I3= 0 2I1 3I2 +0I3 = 4 0I1+3I2+5I3 = 2 I1= 1.226A I2= 0.516A I3= 0.71A b. V= VB1 I1R V= 6 1.226 2 =3.548V 6. A current of intensity 0.5 ampere passes in a solenoid which consists of 20 turns

in each 1 cm, another wire was wound around its mid-point to make one circular turn only its radius is 1 cm. What will be the intensity of the electric current passing in this turn such that its magnetic flux at its center cancels the flux of the solenoid? Explain what happens at the same point if the direction of the current in this turn is reversed. (20A, 2.51 x10-3 Tesla)  Answer BS = BC BT = = + I=20A

= 2.51 x10-3 T

BT = B1+B2 -7

2

7. A uniform insulated metallic wire of cross sectional area 4.25×10 m is wound

tightly as perfectly adjacent turns of one layer around an iron cylinder of diameter cm to form a solenoid. The solenoid then is connected to a battery of EMF (VB = 10 V) and negligible internal resistance. The current passing through the solenoid is 5A. (The resistivity of wire material is 1.7 ×10-8 Ω.m. and the magnetic permeability of iron is 0.002 Wb/A.m.) (π = 22/7) Calculate: a. Number of turns of the solenoid. b. The magnetic flux density at a point along the axis of solenoid.

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 Answer a. I =

b. A = 4.25 r = 3.68

5= R= 2  L=

m =

0.368 m

= L = 50 m N=

B= B = 13.59 T

=

= 500 turn 8. Two parallel wires , the distance between them is 6 cm in the first wire a current of intensity 2 ampere passes in the second wire a current of intensity 1 amperes passes, Calculate: a. The magnetic flux density at a point in middle the distance between them when the direction of current in the two wires at the same direction (2×10-5 T) b. The position of the neutral point when the direction of current in the two wires at the same direction (0.04 m from 2nd wire)  Answer a. b.

BT = = =2

T d1=12-2d1 3d1 = 12 d1 = 4 cm

9. A rectangular loop (12 cm, 10 cm) of 40 turns carrying a current of 2A.

Calculate the torque acting on the loop if it’s placed in a magnetic field density 0.25 Tesla when. a. The plane of the coil is perpendicular to the magnetic flux lines.(0) b. The plane of the coil makes an angle 60 with the magnetic flux.(0.12 N.m) 6 | Page (Revision)

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 Answer a. 0 b. The plane of the coil makes an angle 60 with the magnetic flux.(0.12 N.m) (

=

)

10. A sensitive galvanometer the resistance of its coil is 5 ohms .it gives full scale

deflection when a current of 0.5 milli ampere passes in it .A resistance of 5 ohms was connected to it in parallel such that they formed together one instrument ,then a resistance of 1000 ohms was connected in series with it . It is then used to measure the potential difference. What will be the maximum potential which it can measure. (1.0025v) I = 1 10-3A  Answer Vm = IRm -3 Vm = 1 10-3 1000=1V 0.5 10 5 V = Vg + Vm 2.5 10-3V = 2.5 10-3 + 1 -3 Is = = 0.5 10 A V = 1.0025V I = Is+ Ig I = 0.5 10-3 + 0.5 10-3 11. If the e.m.f of A.C can be calculated from the relation: e.m.f = 100 Sin 9000t, Calculate :a. The effective e.m.f. (70.71 V) b. Value of the e.m.f after 5x10-3 sec. starting from the position normal to field.

(70.71V)  Answer a. e.m.feff = √

= √ b. e.mf = 100 sin 9000t e.m.f = 100 Sin 9000 5x10-3 =70.71V

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12. A step-down transformer of efficiency 80% and the number of turns of its

primary coil is 4000 turns, connected to the end of A.C. power lines to step down the voltage from 3000 volts to 120 volts. If the electric power produced from the transformer is 15 kilowatt. Calculate: a. The number of turns of the secondary coil. (200) b. The current intensity in each of primary and secondary coils.

(6.25 A, 125 A)  Answer a.

Is = 125A

Ns = 200 turns b.

IP = 6.25A

(PW)s= IsVs 15 103 = Is 120 13. A coil of an AC dynamo consists of 500 turns each of cross sectional area 100 cm2 revolves at rate 1500 revolution/min. in a uniform magnetic flux of density 4.2×10-3T. (use ) .calculate: a. The e.m.f. induced when the coil plane makes an angle 60˚ to the field direction. (1.65V) b. The e.m.f. after 0.02 seconds from the parallel position to the field.(6.6V)  Answer a. e.m.f = NBA2 sinθ e.m.f =500 sin30 e.m.f = 1.65V b. e.m.f = NBA2 sin (360ft-90) e.m.f= 500 sin (360 ) e.m.f = 3.3V

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14. When an inductive coil is connected to a D.C supply of 6V and of internal

resistance 1, a current of 1.5 A flows. When the source is replaced by an A.C. supply of 5V and of frequency 49 Hz the current flowing in the same coil is 1A. Calculate the self-inductance of the coil. ( H)  Answer Z= √ I= 1.5 = R= 3Ω V=IZ 5=1 Z Z= 5Ω

5=√ XL = 4Ω XL= 2πfL 4= 2( ) 49 L

L= H 15. As the incident photons energy increases by 20%, the kinetic energy of the emitted photons from the metal surface increases from 0.5 eV to 0.8 eV, calculate the work function for that metal. (Knowing that : = 1 .6 x 10-19 C) K.E2 = 1.2 K.E1 1.2 K.E1 K.E1= (0.8 0.5) 1 .6 K.E1 = E1 EW x 10-19 K.E1 = 2.4 x 10-19 J EW = E1 K.E1 …….(1) By substitute in (1) K.E2 = E2 EW EW = E2 K.E2 …….(1) EW = 0.5 1 .6 x 10-19 – 2.4 x 10-19 From (1) , (2) EW = 1 .6 x 10-19 E1 K.E1= E2 K.E2 K.E2 K.E1= E2 E1

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