29. The eleetrie field in a region of space vae dine -(1.25 NAC) sis 50 where e
ID: 1732079 • Letter: 2
Question
29. The eleetrie field in a region of space vae dine -(1.25 NAC) sis 50 where e is in seconss. The manimm dispincemens perpendicular so & is awre ) 3.32 nA D) 604 nA 13.3 nA 30. A200um coil rotates in a magnetic field of masgnitudes 045 T at a frequency ot se The area of the coil is 5.0 cm. What is the A) 17 v B) 9.4 v C) 2.5 V D) 24 V E) 9.0 v what ist 31. An AC generator su pplies 22 rms volts to a 40-? resistor at 50 Hz. maximum current in the resistor? A) 0.78 A ?) 0.52 A C) 1.0 A D) 21 mA E) 3.3 mA 32. What is the frequency of the light with wavelength 455-nm? A) 5.40 x 1014 Hz B) 6.60 x 104 Hz C) 5.40 x 1017 Hz D) 5.40 x 1015 Hz E) 16.7 kHz 33. The reactance of an inductor is 10 S2 at 2 kHz. The inductance A) 1.59 mH B) 0.80 mH C) 3.45 mlH D) 2.06 mH E) 1.00 mHExplanation / Answer
29 .A
Given electric field is
E = (1.25 N/C) Sin 1500t
Area A = 1.00 m^2
the displacement current perpendicular to the field is I_d = ?
we know that the displacement current is I_d = epsilon0*E*A
I_d = 8.854*10^-12*1.25*1500*1 A = 1.660125*10^-8 A
I_d = 16.6 nA
answer is option A
30. A
N = 200 turns
magnetic field B = 0.45 T , frequency f = 60 Hz
Area A = 5.0 cm^2 =5*10^-4 m^2
maximum emf in the coil e = -N d(phi)/dt
phi is magnetic flux in the coil = B*A cos theta
and theta = w*t , w= 2pi*f
e = N*B*A*wsin wt
e = 200*0.45*5*10^-4*2pi*60*sin(90)
e = 16.964 V
e = 17 V <<<<<<<<----ANS
answer is option A 17 V
31 . A
Ac generator with v= 22 rms voltage
R = 40 ohm
f = 50 Hz
Imax = ?
we have relation between power dissipation , Vrms and resistance is
P_avg = V_rms^2/R = I_max^2*R/2
I_max^2 = 2*V_rms^2/R^2
I_max^2 = 2*22^2/40^2
I_max = 0.7781 A
I_max = 0.78 A
answer is option A
32. B
Given Wavelength Lambda = 455 nm
frequency f = ?
v = lambda*f
f = V/lambda
f = (3*10^8)/455*10^-9) Hz
f = 6.59341*10^14 Hz
f = 6.6*10^14 Hz
answer is option B
33. B
Reactace of inductor XL = 10 ohm at f = 2 kHz
inductance L = ?
XL = W*L = 2pi*f*L = 10 ohm
L = 10/2pi*f
L = 10/(2*pi*2000) H
L = 0.0007957747155 H
L = 0.80 mH
the answer is option B
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