Need a help from 7 through 12 7. A square coil is in a uniform magnetic field th

Question

Need a help from 7 through 12

7. A square coil is in a uniform magnetic field that is perpendicular to the plane of the coil and directed into the screen, as shown. If this magnetic field begins to increase in magnitude what will the induced current in the coil be? a. directed into the screen b. zero C. clockwise d. counter-clockwise e. directed out of the screen 8. A 0.20 H inductor is located in a branch of some circuit that results in a current I through the inductor that passes from A to B, as shown. If 2 volts is produced across the inductor with end A at a lower voltage than end B, what can be said about the current a. It is decreasing at the rate 10 A/s. 0.20 H b. It is decreasing at the rate 0.10 A/s c. It is increasing at the rate 10 A/s. d. It is increasing at the rate 0.10 A/s e. It is equal to 10 A (constant) 9. The switch is closed at t 0. Just after S is closed, what is the voltage across the inductor? a. 5.0 V b. 10 V C. Zero 20 25mH d. 4.0 V e. 20 V 5 10. In the circuit above (#9), what is the current 5 ms after the switch S is closed? a. 1.5 A b. 3.8 A C. 2.5 A d. 0.16 A 63 A 11. An AC generator consists of 6 turns of wire. Each turn has an area of 0.040 m2. The loop rotates in a uniform field (B 0.20 T) at a constant frequency of 50 Hz. What is the maximum induced emf? b. 2.4 V a. 13 V C. 3.0 V d. 15 V e. 4.8 V 12. What is the rms voltage drop across the inductor in the circuit below? b. 27.5 V C. 33 V d. 38.5 V e. 30.5 V in 500 ty

Explanation / Answer

7) B induced will be out of the plane
so I will be ccw so d)

8) V = - L dI/dt

so dI/dt = -V/L = -2/0.2= - 10

so a)

9) all voltage will be across the inductor so 20 V so e)

10)

I = V/R ( 1-e^(-t/tau))

tau = L/R

I = 4*(1-e^(-5.0E-3/(5.0E-3))= 2.5 A so c)

11)

emf = N B A w = 6*0.2*0.04*2*pi*50= 15 V so d)

12) Z = sqrt( R^2 + (w L - 1/( w C))^2)

= sqrt(1000^2 + (500*1 - 1/(500*1.0E-6))^2)= 1803 ohms

Imax = V/Z = 140/1803

V inductor rms = I max w L/sqrt(2) = (140/1803)*500*1/sqrt(2)= 27.5 V

so b)

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