A 1270 turn circular coil has a diameter of 3.84 cm. a length of 12.2 cm, and a
ID: 1566522 • Letter: A
Question
A 1270 turn circular coil has a diameter of 3.84 cm. a length of 12.2 cm, and a resistance of 14.6 ohms. This coil is placed inside a larger 821 turn circular coil. The axes of both coils are co-linear. The diameter of the second coil is 4.86 cm, the length is 14.4 cm, and the resistance is 7.9 ohms. An oscillating voltage is applied to the outer (larger) coil. The voltage applied to the outer coil is varies linearly between V_max = 27 V_max and = -27 V. The period (T) of one complete cycle of this oscillation is 33.1 ms. Since the Voltage varies linearly, the graph of the voltage vs. time makes a triangular wave form as show below. What is the maximum value of the current through the outer coil What is the What is the What is the time of one cycle of oscillation of the magnetic field through the outer coil What is the maximum value of the magnetic flux through the inner coil Tm What is the What is the What is the time of one cycle of oscillation of the voltage across the inner coil What is the maximum value of the induced current through the inner coil The graph of the induced voltage across the inner coil vs. time has the form of a wave.Explanation / Answer
let
N1 = 1270 turns
d1 = 3.84 cm
L1 = 12.2 cm
R1 = 14.6 ohms
N2 = 821 turns
d2 4.86 cm
L2 = 14.4 cm
R2 = 7.9 ohms
1) The maximum current through outer coil, Imax2 = Vmax2/R2
= 27/7.9
= 3.42 A
2) same as voltage.
T = 33.1 ms
3) Bmax2 = mue*N2*Imax2/L2
= 4*pi*10^-7*821*3.42/0.144
= 0.0245 T
= 24.5 mT
4) same as Current.
T = 33.1 ms
5) maximum value of the magnetic flux through the inner coil = A1*N1*Bmax2
= (pi*d1^2/4)*N1*Bmax2
= (pi*0.0384^2/4)*1270*24.5*10^-3
= 0.0360 T.m^2
6) T = 33.1 ms
7) maximum induced emf in the inner coil, Vmax1 = change in magnetic/time taken
= (0.036 - (-0.036))/(T/2)
= 0.036*2/(33.1*10^-3/2)
= 4.35 V
8) T = 33.1 ms
9) Imax1 = Vmax1/R1
= 4.35/14.6
= 0.298 A
10) square wave
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