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Topic 5 , 10-11 New Selected Problems 2 - Solutions 1. a) If the distance from a point charge is doubled the magnitude of the electric field will _______________________________. b) The electric field of a point charge at distance r is 1.0 x 10-4 NC-1 . If the distance r is reduced to one- third , what is the magnitude of the new electric field? 2) An isolated electron is acted on by an electric force of 3.2 x 10-14 N. What is the magnitude of the electric field at the electron’s location? Answers : 1. a) If the distance from a point charge is doubled the magnitude of the electric field will decrease by one fourth E=k Q r2 1 (2)2 =1 4 b) The electric field of a point charge at distance r is 1.0 x 10-4 NC-1 . If the distance r is reduced to one- third , what is the magnitude of the new electric field? E= 1 = r2 1 . = (1/3)2 9 so it is 9 times original field = 9.0 x 10-4 NC-1 1 2. An isolated electron is acted on by an electric force of 3.2 x 10-14 N. What is the magnitude of the electric field at the electron’s location? Answer : no distance given so use E = F q electron = 1.6 x 10-19 C E = 3.2 x 10-14 N 1.6 x 10-19 C = 2.0 x 105 NC-1 ( 2 sig. dig.) 3. There are two separate series circuits using a 12 V battery and one resistor. Which circuit consumes more power : a circuit with a 5.0 -Ω resistor or a 10 -Ω resistor ? Explain. P is proportional to current. The 5.0 -Ω resistor circuit will have less resistance and therefore more current and consume more power. Can Calculate using both formulas : P = RI 2 ( need to calculate I from V = RI ) P = I2 R ( 5 ohm resistor) calculate current first : P = V2 R I = V/R = 12V/ 5 = 2.4 A P = ( 2.4)2 x 5 = 28.8 W P = I2 R ( 10 ohm resistor) calculate current first : I = V/R = 12V/ 10 = 1.2 A P = ( 1.2)2 x 5 = 14.4 W P = V2 R = ( 12) 2 = 28.8 W 5 P = V2 R = ( 12) 2 = 14.4W 10 power is inversely proportional to R 2 4. In the circuit below, the voltmeter has a resistance 100 k. The battery has negligible internal resistance and emf 6 V. The reading on the voltmeter is B A. 0 V. B. 2 V. C. 3 V. D. 4 V. 3 5. OMITTED 4 6. 5 6Answer C Direction is from Y X. Y is + and X is of + charge. . Direction follows movement In the case of parallel plates the electric field is determined by E = V/ d where d is the distance between the plates. The electric field is uniform and has the same value at all points between the plates. Its direction is from high potential to low potential. fig. 2.15 p. 294 Work done to move a charge is not uniform and depends on distance the charge moves : W = Eqd = qV 7. OMITTED 6 8. 7 8. 8 Ohm’s law Review Normal Ohm’s law – ohmic resistor Filament = NON OHMIC resistor that heats up and this heat will increase the resistance exponentially NOT linearly as in normal ohm’s law. This increase in resistance will cause a DECREASE in current ( I ) as V increases. V ___________________ I Note if the graph is inverted to I vs V it will look like this I V 9 Ohm’s Law Lab Review Variable resistor also called a potentiometer. It varies the resistance and thus controls the current. Make sure you can relate the lab equipment to the circuit diagram below it. Be able to label different parts of the circuit. Be able to label the position of the potentiometer to give you full resistance and full voltage drop or zero resistance and zero voltage drop. The more of the potentiometer you use the more the resistance and the more the voltage drop. If the arrow coming from the ammeter is placed closest to the positive pole of the potentiometer then you will get zero resistance and zero voltage drop. If the arrow coming from the ammeter is placed furthest from the positive pole of the potentiometer then you will get full resistance and full voltage drop. 10 X 11 9-12 HL only 9. A circular coil of wire of radius r is placed in a uniform magnetic field of flux density B. The angle between the plane of the coil and the magnetic field is θ. B r The magnetic flux linking the coil is B A. πr2B. B. πr2B sin θ. C. πr2B θ. D. πr2 NOTES REVIEW Ф = BA Area ( A) = πr2 Magnetic Flux ( Ф) , Faraday´s Law Ф = BA Since the induced emf or voltage depends on magnetic flux or field lines, magnetic flux (Ф) is defined as the number of field lines in a magnetic field ( B) that pass through a loop of area (A) : Ф = BA units: magnetic field B = T ( telsa Area (A ) = m2 Magnetic flux (Ф ) = BA = T m2 ( weber Wb) 12 MAGNETIC FLUX AT DIFFERENT ANGLES Figs. 20. 4 b, c d and e below show how the loop angle ( θ) with respect to the magnetic field (B) can change the value of the magnetic flux ( Ф) . Note: Magnetic flux is MAX at θ = 0 or 180…perpendicular formation Fig.b,c Magnetic flux is ZERO at θ = 90 …..parallel formation 13 10. The magnetic flux Φ through a coil having 500 turns varies with time t as shown below. 2.5 2.0 1.5 –3 / 10 Wb 1.0 0.5 0.0 0 1 3 2 –3 t / 10 s 4 5 The magnitude of the emf induced in the coil is C A. 0.25 V. B. 0.50 V. C. 250 V. D. 1 000 V. FARADAY`S LAW and N : ε=NΔФ t ε=NΔФ t = 500 x 2 x 10-3 Wb 4 x 10-3 s induced emf or voltage (V) with # of loops N = 250V note can choose any 2 values for Ф and t off straight line 14 Notes Review FARADAY’S LAW ε= ΔФ t induced emf or voltage (V) With these definitions for flux and flux linkage, we can now write the results of Faraday’s investigations. N B t (Faraday's Law) FARADAY`S LAW and N : ε=NΔФ t induced emf or voltage (V) with # of loops N States that the induced emf also depends on the number of loops (N) in a coil along with the time rate change in magnetic flux . SUMMARY of conditions needed to induce an emf ( voltage and subsequent current) . Only one of the following needs to be met : 1. The strength of the magnetic field ( B) changes 2. The loop area ( A) changes 3. The angle between the loop and the field changes Key word: change = fluctuates = flux 15 11. This question is about electrical energy and associated phenomena. electromagnet The current in the circuit is switched on. (i) a) State Faraday’s law of electromagnetic induction and b) use the law to explain why an emf is induced in the coil of the electromagnet. Answer a) : (induced) e.m.f. proportional to rate of change of magnetic flux (linkage); FARADAY’S LAW ε= ΔФ induced emf or voltage (V) t States that the induced emf or voltage depends on the change in the magnetic flux over time. Answer b) : as current increases, magnetic field in coil increases and thus change in flux and e.m.f. is induced NOTES REVIEW States that the induced emf also depends on the number of loops (N) in a coil along with the time rate change in magnetic flux . SUMMARY of conditions needed to induce an emf ( voltage and subsequent current) . Only one of the following needs to be met : 1. The strength of the magnetic field ( B) changes 2. The loop area ( A) changes 3. The angle between the loop and the field changes Key word: change = fluctuates = flux 16 (ii) State Lenz’s law and use the law to predict the direction of the induced emf in (i). Answer: induced e.m.f. must oppose e.m.f. of battery LENZ´S LAW : Induced field Opposes the External-Original field to resist any change in magnetic flux Due to conservation of energy and action – reaction forces : LENZ´S LAW states that when a magnetic field induces an emf in a loop (called the external - original field) ; the loop in turn, generates a secondary magnetic field (called the induced field )that is OPPOSITE in direction of the original magnetic field that caused the induction: Newton´s 3rd Law: Axn - Rxn Same as Rt. Hand rule except 2 magnetic fields are involved and you have to find the INDUCED Field fig. B above NOT THE EXTERNAL FIELD fig. a above. 17 12. 18 12. Answer : (i) Calculate the rms value of the emf of the generator. From graph : peak reading = 1400 ± 50 V peak reading 990 V 2 Vrms = (ii) The speed of rotation of the generator is halved with no other changes being made. On the graph, sketch the variation of emf with time. Voltage is directly proportional to speed of rotation. Speed is halved so voltage is halved giving peak of 700V: 1400v/ 2 = 700 V (iii) Explain why the graph you drew in (ii) is different from the original graph. Voltage is directly proportional to speed of rotation. Speed is halved so voltage is halved giving peak of 700V: 1400v/ 2 = 700 V 19 13. Two long, vertical wires X and Y carry currents in the same direction and pass through a horizontal sheet of card. X Y Iron filings are scattered on the card. Which one of the following diagrams best shows the pattern formed by the iron filings? (The dots show where the wires X and Y enter the card.) A. B. C. D. Answer = A C is close but does not show circular magnetic field produced by straight wire. 20