*Please just see pictures attached and answers the problems posted in the pictur
ID: 2299812 • Letter: #
Question
*Please just see pictures attached and answers the problems posted in the pictures. Make sure to answer all parts of the question.
1. Consider an ideal, infinitely long solenoid that carries a current of 15 mA, has a wire loop
density of 5000 turns/meter, and is 5 cm in diameter
2. A straight wire and a wire loop are arranged as shown below. Both are insulated so they are not in electrical contact with each other. A current of 10 Amps flows through to the right in the horizontal wire. For times ??>0, the current decreases according to ??=10???????(2????????)??. For times ??>0 what emf is induced in the loop, including the direction?
1. Consider an ideal, infinitely long solenoid that carries a current of 15 mA, has a wire loop density of 5000 turns/meter, and is 5 cm in diameter. A) What is the magnetic field inside the solenoid? Answer in vector notation in Tesla using the axes shown B) For the isosceles right triangular Amperian loop shown, calculate fB dš. Answer in Tesla- meter C) For the isosceles right triangular Amperian loop shown, calculate ui through the loop. Answer in Tesla-meter 5 cm 1 cm ty l cmExplanation / Answer
1. I = .015A, n=5000 turns/m, mu_0=4pi * 10^-7
a. B=mu_0 * n * I
=(4pi * 10^-7) * (5000) * (.015)
= 9.425 * 10^-5 T (-j)
b. int(ds)= 12cm - 2cm = .10m (-j)
B*int(ds)=(-9.425 * 10^-5)(-.10)
= 9.425 * 10^-6 Tm
c. mu_0*i = (4pi * 10^-7)(.12m(5000/m)-.02(5000/m))(.015)
= 9.425 * 10^-6 Tm
2. MagFlux = int(BdA), emf=-d/dt(MagFlux)
MagFlux = int( [mu_0 * I / (2pi*r)] * L * dr)
= [mu_0 * I * L / (2pi*r)] * {[int(1/r dr),r=0 to .1m] - [int(1/r dr),r=0 to .04m]}
= (2 * 10^-7)(.07m)(I) * [int(1/r dr),r=.04 to .1m]
= (2*10^-7)(.07)(ln .1 - ln .04) * I
= (1.283 * 10^-8)* (10-2t)
= 1.283*10^-7 - (2.566*10^-8)t
emf = 2.566*10^-8 V (just derivative of MagFlux w/respect to t)
emf induced CCW, by RHR
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