Practice Problem 21.9 SOLUTION SET UP AND SOLVE At time t = 3.0s, the current in
ID: 1507516 • Letter: P
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Practice Problem 21.9 SOLUTION SET UP AND SOLVE At time t = 3.0s, the current in coil 2 is A Tesla coil is a type of high-voltage generator that was invented by Nikola Tesla (1856-1943). In one form of Tesla coil, shown in (Figure 1), a long solenoid with length l and cross-sectional area A is closely wound with Ni turns of wire. A coil with N2 turns surrounds it at its center. You have a Tesla coil with 1 0.50m. A-10cm2 N; = 1000 turns, and N2-10 turns. Let's apply a time- varying current to the blue coil. Suppose the current i2 is given by 12 (2.0 × 106A/s)t. At time t = 3.0. what is the average magnetic flux through each turn of the solenoid caused by the current in the smaller coil? What is the induced emf in the solenoid? iy = (2.0 x 106A/s)(3.0 x 10 %) = 6.0A To find the flux in the solenoid, we need to know the mutual inductance of the coils AoAN1N2 _ (41TX 10-7 Wb/(A-m))(10x10-4 m2)(1000)(10) M- 0.50m 2.5× 10 5H Now we can solve Ni m-M221 for the average magnetic flux in coil 1 (2.5x10-oH)(6.0A) 1000 Mi 1.5×10 7Wb This is an average value; the flux will vary considerably from the center to the ends The induced emf. E1 , is given by -M22/ with the change in current per unit time. 22/M. equal to 2.0×106 A/s Figure 1 |ofl REFLECT In an operating Tesla coil, 22/t would be alternating much more rapidly, and its magnitude would be much larger than in this example Cross-sectional area A Part A Practice Problem: For the given Tesla coil, how many tums (Ni) should coil 1 have in order to get an induced emf magnitude of 350 V Express your answer in turns to two significant figures. Blue coil: N turns Black coil: Ni turnsExplanation / Answer
N
umber of turns in inner solenoid = N1 = 1000 turns
Number of turns in outer solenoid = N2 = 10
Length of inner solenoid = L = 0.50 m
The current in the inner solenoid = i = 2 * 10^6 A
Radius of solenoid = r=diameter/2= 2.00/2=1.00 cm=0.01 m
Rate of increasing of current =di/dt = 1800 A/s.
The average magnetic flux through each turn of the inner solenoid. =B.A=(uo)(N/L)I.A
The average magnetic flux through each turn of the inner solenoid. =[4(pi)*10^-7](1000/0.50)(2 ×10^6)(0.01)
A) The average magnetic flux through each turn of the inner solenoid. =50.275 Weber
b) Induced EMF = (uo) N^2 × L/A
= 4(pi)(10^-7) * 0.50/0.01
= 0.628*10^-4 V
(51)
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