first. As 22. A bar magnet is falling through a loop of wire with constant veloc
ID: 2039232 • Letter: F
Question
first. As 22. A bar magnet is falling through a loop of wire with constant velocity. The north pole enters the south pole leaves the loop of wire, the induced current (as viewed from above) will be b. counterclockwise. c. zero d. along the length of the magnet. A bar magnet is falling through a loop of wire with constant velocity. The south pole enters first. As the magnet leaves the wire, the induced current (as viewed from above): a. is clockwise. b. is counterclockwise. c. is zero. d. is along the length of the magnet. 24. A lat coil of wire consisting of 15 turns, each with an area of 90 cm2, is positioned perpendicularly to 5.9T in 2.0s. I a uniform magnetic field that increases its magnitude at a constant rate from 3.5T to the coil has a total resistance of 0.20?, what is the magnitude of the induced current? a. 1990 mA b. 81 mA c. 810 mA d. 130 mA e. 320mA 25. A flat coil of wire consisting of 20 turns, each with an area of 80cm2, is positioned perpendicularly to a uniform magnetic field that increases its magnitude at a constant rate from 2.0T to 6.0T in 20s. If the coil has a total resistance of 0.40 ?, what is the magnitude of the induced current? a. 500 mA b. 70 mA c. 140 mA d. 800 mA 26, An AC series circuit contains a res stor of 100, a capacitor of 1.01 ?F and an inductor of 116mH. What frequency should be used to create a resonance condition in the cireuit? a 147 Hz b. 1360 Hz e. 1170 Ha d. 465 Hz e. 930HzExplanation / Answer
22) b
23) a
24) N = 15 , A = 90cm^2, B1 = 3.5 T, B2 = 5.9 t , t = 2s , R = 0.2ohms
Induced emf V= d(phi)/dt = NA (B2 -B1)/t
V = IR
15*90*10^-4*(5.9-3.5)/2 = I*0.2
I = 0.81 A
I =810 mA
correct option is (c)
25) N = 20 , A = 80cm^2, B1 = 2 T, B2 = 6 t , t = 2s , R = 0.4 ohms
Induced emf V= d(phi)/dt = NA (B2 -B1)/t
V = IR
20*80*10^-4*(6-2)/2 = I*0.4
I = 0.8 A
I =800 mA
correct option is (d)
26) L = 116 mH, C = 1.01 uF
f = (1/2pi)(1/LC)^0.5
f = (1/6.28)(1/(0.116*1.01*10^-6))^0.5
f = 465 Hz
correct option is (d)
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