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An 820 -turn wire coil of resistance 24.0$\Omega$ is placed on top of a 12500 -turn, $7.00-\mathrm{cm}$ -long solenoid, as in Figure $\mathrm{P} 20.57 .$ Both coil and solenoid have cross-sectional areas of $1.00 \times 10^{-4} \mathrm{m}^{2}$ (a) How long does it take the solenoid current to reach 0.632 times its maximum value? (b) Determine the average back emf caused by the self-inductance of the solenoid during this interval. The magnetic field produced by the solenoid at the location of the coil is one-half as strong as the field at the center of the solenoid. (c) Determine the average rate of change in magnetic flux through each turn of the coil during the stated interval. (d) Find the magnitude of the average induced current in the coil.

a. 20ms

b. 37.9 V

c. 1.52 mV

d. 51.8 mA

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a number of tons capital and equals to 1 to 5 double zero and the length of the solenoid capital l equals to seven centimetres and area of the solenoid is gone into 10 to the power minus 4 m square. So the self indulgence of the solenoid capital is given by a new note capital and the square A by help substituting the values. In this equation, we get eloquence communities for primary perfect and to the par minus seven and N is 1 to 5, Double zero Holy square area is going to turn to the par minus four metre square and l A seven centimeter, that is 7 to 10 to the minus 2 m. So from here all comes out to be 0.20 Henry, this is our part. Okay? So no, the time constant for RL circuit is given by how equals two l by our and L H 0.20 and R s 14 home. So from here comes out to be 20 milliseconds. So this is the answer for the part of the problem. Okay? No moving to the part B in which we have to calculate the back e m f. Okay, so Delta I even b equals two Delta. We buy our and the Delta V is 60 world and R s 14 home. So from here, Delta comes out to be 4.29 MPs. Okay, so in one time, constant that I will be equals two. 0.632 I okay? And I is actually Delta I. So from here we get 0.632 multiplied by 4.29 MPs. So from here, I doubt comes out to be 2.71 emptier. Okay, So back e m f e l will be equals to El Delta. I and they were way Delta. So delta l is 0.280 and Delta I will be this value. So 2.71 divided by Delta T is 20 milliseconds. Okay. And one time constant. So from here, after solving e l comes out to be 37.9. Vote. This is the answer for the part of the problem. Okay, No moving to the part C in which we have to calculate the changing magnetic flux. So changing magnetic flux. Adele Fiber dlt is equals two daily Peabody lt multiplied by area. Okay. And it can be written as that they'll be is equals to Dale B as by two that is magnetic field produced by the solenoid. At the location of the coil is one half of the magnetic field produced at the center of solenoid. So from here we get Delphi. By Delta P equals 21 by two Delta BS. They were by Delta T manipulated by area. Okay. And Delta Bs will be equals two one by two. New note. And they were by two. Well, Delta High by Delta t Molecular by area. So now substituting values, we get Delta fiber Delta T equals 21 by two Molecular by communities 4.2 10 to the power minus seven and n is 1 to 5 double zero. They were by two into l l. A. Seven centimeter, that is 7 to 10 to the four minus two. And delta is group and seven when M. Pierre and Delta T is 20 milliseconds okay, and area is one into 10 to the power minus 4 m square. So from here, after solving, we get Delta five by Delta t equals to 1.52 million volts. So this is the answer for the party of the problem. Okay, No moving to the party in which we have to calculate the current I do. So we can write. I current will be equals two e two by our two. And it will be equals two and two Delta fiber Delta T. So, delta fiber delta T this value is obtained in the party as this. So from here, we can write, I, too will be equals two and two is 8 20 Delta five to Delta Fiber. Delta Delta T is 1.52 but their weight into the four minus three. Okay. And they were by 24. Okay, so from here, I comes out to be 51.8 million.