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A square, single-turn wire loop $\ell=1.00 \mathrm{cm}$ on a side is placed inside a solenoid that has a circular cross section of radius $r=$ $3.00 \mathrm{cm},$ as shown in the end view of Figure $\mathrm{P} 20.18$ . The solenoid is 20.0 cm long and wound with 100 turns of wire. (a) If the current in the solenoid is 3.00 A, what is the flux through the square loop? (b) If the current in the solenoid is reduced to zero in 3.00 s, what is the magnitude of the average induced emf in the square loop?

a. =1.39 \times 10^{-1} \mathrm{T} \cdot \mathrm{m}^{2}

b. =3.475 * 10^{\wedge}-8 \mathrm{volts}

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Numerade Educator

University of Sheffield

for part A were asked to find the flux through the swear loop. If the current in the solenoid is three amperes so we can find the magnetic field first, then find the flux. So the magnetic field be is equal to you, not which is just the magnetic permeability of freeze face, constant times, the number of turns times the current, which we're told is three amperes divided by the length, which is 0.2 meters. So plugging all those values into this expression, we find that this is equal to 1.88 times 10 to the minus three, then units for magnetic field or Tesla. Therefore, the magnetic flux slot fine is going to be equal to the magnetic field we just found, multiplied by the area of the square loop, which is 10 to the minus four square meters times the co sign of the angle data. Well, since the angle fate is the angle between the magnetic field on the plane of the loop and that zero degrees co sign of zero is one. So this is just the magnetic magnetic field times the area, which comes out to be 1.8 times 10 to the minus seven. Mrs. Tesla, Times meter squared again. That's because Veda was equal to zero degrees. Okay, so for part B were asked to find the magnitude of the induced E M f after three seconds. So the magnitude is going to be equal to the magnitude of the change in fi. But five is going from the five, uh, of the magnetic flux we just found, which is 1.8 attempts and minus 7 to 0. So that changes just the magnetic flux we just found. Divided by the change in time, which we're told it takes three seconds to go from that value. We just found in part A to zero. So playing three seconds in for Delta T in the value we just found for Delta Phi for the change of magnetic flux, we find that this is equal to 6.27 times 10 to the minus eight, and the units here are bolts making box that it is our solution for part B