Hello Everyone!. I have an exam coming up in a few days and I wanted to see how

Question

Hello Everyone!. I have an exam coming up in a few days and I wanted to see how someone else answers these questions. It is from a college physics 2 class. Electrical physics questions. If you guys want to answer these and show your work and any equations you use it would be helpful. I can get through most of them but it would be helpful to see how other people also get through them. The answer key is also here so that should help you know you get the right answer. Have fun. I have another question like this i just posted so if you want to take a shot at that one after this then thats cool. It is similar stuff so if this easy then it is an easy way for you to double your points. Quality of answers will get the points not speed.

Answer Key:

1: 5

2: 1

3: 5

4: 5

5: 1

6: 4

7: 3

8: 2

9: 3

10: 2

11: 1

12: 5

13: 5

14: 4

15: 4

Mu 0 = 4 pi times -7 T m/A Speed of light = 3 times 108 m/sec A simple loop wire that carries I to the right... Under which circumstance would a clockwise current be induce in the loop? Move the loop right. Move the loop left. Decrease I. Move the wire down. Move the loop down. The magnetic field between the poles of the magnet below is 10 Wb/m2. A loop 0.3 m tall and 0.2 wide is placed between the poles. What is the magnetic flux through the loop? Zero 0.6 Wb 2 Wb 3 Wb 6 Wb A square coil, enclosing an area of 4 times 10-4 m2, is wrapped with 5000 turns of wire and has a resistance of 0.4 ohm. A uniform magnetic field points into the page and increases from zero to 0.25 T during an interval of 1.0 s. What current is induced in the coil? 0.125 Amps counterclockwise. 0.5 Amps clockwise. 0.5 Amps counterclockwise. 1.25 Amps clockwise. 1.25 Amps counterclockwise. A 0.2 m long metal rod moves along two frictionless, conducting rails at a constant speed of 5 m/s. The rails have negligible resistance, but the rod has a resistance of 2 Ohms. Zero 2 Amps clockwise. 3 Amps clockwise. 3 Amps counterclockwise A motor with a coil resistance of Ohm is attached to a voltage supply of 80V. What is the current when the motor is running at its rated speed with a back emf of 70 V? 1 A 7 A 8 A 15 A Zero A transformer is to be designated to increase the 30 kV (rms) output of a generate to the transmission line voltage of 345 kV(rms). If the primary winding has 80 turns, how many turns must the secondary have? 6 70 700 920 9200 A long time after the switch has been closed, what energy is stored in the circuit below? 0.1 Jules 0.2 Joules 0.5 Joules 1.0 Joules 50 Joules A solenoid of radius 2 cm has 500 turns and a length of 30 cm. What is the induced emf if the current through the coil is changing at a rate of 2 A/sec? 0.84 mV 2.63 mV 4.58 mV 6.33 mV 8.27 mV Two loops are wrapped around an iron cylinder. If the switch on the loop with the battery is suddenly closed, then a current is induced in the secondary loop which: flows from A to B constantly. flows from B to A constantly. flows from A to B for a brief amount of time, flows from B to A a brief amount of time. alternates. A 20 Ohm resistor is connected to a power supply. The voltage of the power supply. The voltage of the power supply is shown in graph. What is the RMS current in the circuit? 0.4 Amps 1.77 Amps 2.50 Amps 3.00 Amps 3.54 Amps A capacitor is connected to the power supply shown below. The rms voltage is 120 Volts and the peak voltage is 170 Volts. What capacitance is needed to keep the maximum current in the circuit below 0.2827 Amps? 4.41 mu F 6.25 mu F 9.12 mu F 15.8 mu F 27.4 mu F Consider a series RC circuit with a capacitance of 44.2 mu F and an unknown resistance R. The circuit is driven by a 120 Volt, 60 Hz AC generator. If the rms current in the circuit is 1.2A, calculate the resistance R. 30 Ohms 40 Ohms 50 Ohms 60 Ohms 80 Ohms An AC series circuit contains a resistor of 20 Ohm, an inductor of 30 mH and a capacitor. If the frequency of the applied voltage is 500 Hz, what should the capacitance be if resonance is achieved. 0.38 mu F 0.81 mu F 1.59 mu F 2.42 mu F 3.38 mu F Graph (A, B) could represent: The (A - voltage, B - current) for a resistor in an ac circuit. The (A - voltage, B - current) for a resistor in an ac circuit. The (A - voltage, B - current) for a capacitor in an ac circuit. The (A - voltage, B - current) for a inductor in an ac circuit. None of the above. Orange - red light has a wavelength of 600 nm. What is the frequency of 600 nm. What is the frequency of this light? 2 times 10-35 Hz 2 times 10-12 Hz 5 times 1011 Hz 5 times 1014 Hz 5 times 1017 Hz

Explanation / Answer

1:5)

The magnetic field through the loop is out of the page and is increasing if we move the loop down. And therefore aclockwise current induces in the loop according to the Lenz's law.

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2:1)

flux = A B cos(theta) = A B cos(90) = 0

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3:5)

emf = d(phi)/dt = N A (dB/dt) = 5000*4e-4*(0.25/1) = 0.5 V

i = emf/R = 0.5/0.4 = 1.25 A

current is counter-clockwise, according to the Lenz's law.

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4:5)

emf = d(phi)/dt = (B L dx)/dt = B L v

i = emf/R = B L v/R = (6*0.2*5)/2 = 3 A

current is counter-clockwise, according to the Lenz's law... as the rod moves to the right, flux increases and according to the Lenz's law a current must be induces counter-clockwise to decrease the flux.

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5:1)

V = 80 - 70 = 10 V

i = V/R = 10/10 = 1 ohms

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6:4)

N2/N1 = V2/V1

N2 = (V2/V1) N1 = (345/30) * 80 = 920 turns

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7:3)

i = V/R = 10/2 = 5 A

energy = 0.5 L i^2 = 0.5 * 0.040 * (5)^2 = 0.5 J

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8:2)

B = u0 N i/L

==> emf = d(phi)/dt = d(N A B)/dt = N A dB/dt = N A u0 N/L (di/dt) =N pi R^2 u0 N/L (di/dt)

==> emf = N pi R^2 u0 N/L (di/dt) = 500*3.1416*0.02*0.02*4*3.1416e-7*500/0.30 * 2 = 2.63 mV

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9:3)

when the switch is closed, a magnetic field is induced in the iron cylinder. the magnetic field is increases and therefore according to the Lenz's law, a magnetic field must be induce to the left. Therefore a current must be from A to B.

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10:2)

V_max = 50 V

i_max = V_max/R = 50/20 = 2.5 A

i_rms = i_max/sqrt(2) = 2.5/sqrt(2) = 1.77 A

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11:1)

i_max = V_max/X

==> X = V_max/i_max = 170/0.2827 = 601.34 ohms

X = 1/(w C) = 1/(2 pi f C)

==> C = 1/(2 pi f X) = 1/(2*3.1416*60*601.34) = 4.41 x 10^-6 F = 4.41 uF

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12:5)

Z = V/i = 120/(1.2) = 100 ohms

X = 1/(2 pi f C) = 1/(2*3.1416*60*44.2e-6) = 60.01 ohms

Z = sqrt(R^2 + X^2)

==> R = sqrt(Z^2 - X^2) = sqrt(100^2 - 60.01^2) = 80 ohms

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13:5)

resonance ==> w = 1/(sqrt(L C))

==> (2 pi f)^2 = 1/(L C)

==> C = 1/(L (2 pi f)^2) = 1/(0.030*(2*3.1416*500)y2) = 3.38 x 10^-6 F = 3.38 uF

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in a ac inductor circuit the voltage wave have a

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