Question

1) The armature of an ac generator is a rectangular coil 2.00 cm by 6.00 cm with 80 turns. It is immersed in a uniform magnetic field of magnitude 0.250 T. If the amplitude of the emf in the coil is 22.0 V, at what angular speed is the armature rotating?

_______rad/s

2) The magnetic field between the poles of an electromagnet is 3.30 T. A coil of wire is placed in this region so that the field is parallel to the axis of the coil. The coil has electrical resistance 24.6 Ω and radius 1.80 cm. When the current supply to the electromagnet is shut off, the total charge that flows through the coil is 9.00 mC. How many turns are there in the coil?

_____turns

3) An external magnetic field is parallel to the central axis of a 65.0-turn coil of radius 7.50 cm. During an interval of 156 ms, the field changes from 0.180 T in one direction to 0.340 T in the opposite direction. The resistance of the coil is 70.0 Ω. What is the average induced current in the coil during this interval?

________ mA

4) Suppose you wanted to use Earth’s magnetic field to make an ac generator at a location where the magnitude of the field is 0.05000 mT. Your coil has 1008 turns and a radius of 3.700 cm. At what angular velocity would you have to rotate it in order to generate an emf of amplitude 1.000 V?

______ rad/s

5) A 3.70-kW heater is designed to be connected to a 120-V rms source. What is the power dissipated by the heater if it is instead connected to a 120-V dc source?

______kW

6) What is the reactance of an air core solenoid of length 6.00 cm, radius 1.20 cm, and 247 turns at a frequency of 21.0 kHz?

_______Ω

7) At what frequency does the maximum current flow through a series RLC circuit containing a resistance of 4.50 Ω, an inductance of 302 mH, and a capacitance of 493 pF?

_____kHz

8) A 22.0-mH inductor, with internal resistance of 22.0 Ω, is connected to a 110-V rms source. If the average power dissipated in the circuit is 62.0 W, what is the frequency? (Model the inductor as an ideal inductor in series with a resistor.)

_____Hz

Answer 1

17 emf induced Ox 0% والا اللہ کا ورد Nded = Nd BÅND BACSO e = NBAD Coswt) = -NBAW Sint amplitule of emf = NBAW. > 22 = 80X0.25x12x10-4 W W = 916.67 W=916.67 rad/sec B = emf induced BA AR at Let This Charge como flows in At time then emf induced omf of the Coil BA e=ndo =NBA At 2) Let at 7 NBA At IR NBA=I AAR > N = Q R BA N = 94163x 24.6 3.34* (18)²x10-4 N= 65.95 turns or N = 66 turns. Ap 37 unduced Current induced voltage, R (N 0%) = NA AB le 65X X X 4 . 5)²x10-4x (0.18 - (- 0:34] 156810²3x70 A = i= 54.71X10-3 A d=54.71mA. R 7 1 =

47 Same as problem (3) amplitude of emp NBAW 1 = 100SX0.05X10-3x3-14X6-739x1074 W W = 4615.69 red/s. W = 4615 69 rad/sec.