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Problem 19 Easy Difficulty

A 300-turn solenoid with a length of 20.0 cm and a radius of 1.50 cm carries a current of 2.00 A. A second coil of four turns is wrapped tightly around this solenoid, so it can be considered to have the same radius as the solenoid. The current in the 300-turn solenoid increases steadily to 5.00 A in 0.900 s. (a) Use Ampère’s law to calculate the initial magnetic field in the middle of the 300-turn solenoid. (b) Calculate the magnetic field of the 300-turn solenoid after 0.900 s. (c) Calculate the area of the 4-turn coil. (d) Calculate the change in the magnetic flux through the 4-turn coil during the same period. (e) Calculate the average induced emf in the 4-turn coil. Is it equal to the instantaneous induced emf? Explain. (f) Why could contributions to the magnetic field by the current in the 4-turn coil be neglected in this calculation?

Answer

a. 0.00377 T
b. 0.00942 T
c. 0.000707 \mathrm{m}^{2}
d. 0.0000160 N . m^{2}
e. 1.78 \times 10^{-5} \mathrm{V}
f. no answer available

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Video Transcript

{'transcript': "for part of our question were asked to find the resistance of the copper wire that we have. Uh, well, resistance of copper wire we call that capital are subsea is going to be equal to the resisted ity of the copper wire. Call that row subsea, which is 1.7 times 10 to the minus eight homes per meter. And that's table 17.1 in the book. If you wantto check that in the book multiplied by the length of the copper wire that we have called El Subsea and then divided by the area of the copper wire, which since it's circular, it's pi times the radius of the copper wire up little are some C square. Plugging those values into this expression, we find that this is equal to 1.3 in the units here, our rooms so we can go ahead and box set. It is their solution for party part B. So the length of a single turn of wire here is going to be equal to the circumference of the wire or two pi R s. So now, to find the number of turns of this wire, we can use the equation for the length of the wire here, which is L subsea, is equal to the number of terrace times circumference, which is to times pi times uh, our suggests the circumference of the soldierly. So now, solving for the number of turns, we find that the number of turns is equal to the length of the copper wire divided by ah two times pi times the radius of this Illinois Plugging those values into this expression, we find that this is equal to 477.46 But you can't have a fraction of a turn here, so it's better to go over or ah, around up here. So this would be rounding off too 400 in 78 turns, but to two significant figures. This is 480 turns. And, uh, since this is approximately and not exactly maybe we should just use approximately equal to 480 turns here using proper notation so we can go ahead and box said it is our solution for part B cool blackout are solution with the box. Okay, so now moving on for part C. Part C has asked us to find the length of the soul annoyed. So to find the length of the solenoid that's just going to be equal to the number of turns in the solenoid multiplied by the diameter of the wire in these turns, right? So each cause you're layering these on top of each other, so the thickness of the wire, the diameter of the wire on top of each other, is going to determine the length of the solenoid. So we'll call this. The multiplied by the diameter of the copper wires will call a decent see, of course, diameters just two times the radius. So this is number of turns times two times the radius of the copper wire complaining those values into this expression. This comes out to equal 0.48 meters. Weaken box set in is their solution for part C party party assets to find the induct in CE. Now, in this, uh, this Illinois, this is going to be equal to induct its Is it denoted by capital L says mu. Not which is the vacuum permitted permitted ity of free space. It's four pi times 10 to the minus seven, but you can look that up if you want to times the number of turns squared times the area divided by the length of this Illinois of course area of the solenoid is just gonna be pi times the radius of the solenoid squared. So this is you, not times the number of turns squared times pi times the radius of the solenoid squared divided by the length of the solenoid which we just found in part c Looking those values into this expression, we find that D is equal to 0.76 times 10 to the minus three. Henry's 10 minus three is Mila. So this is three point our 3.0.76 Millet Henry's making box that in as our solution for R D. Okay, now moving on for part E party has asked to find the time constant for the circuit. So the equation for time constant in a r l circuit, which has a resistor and, uh, a solenoid in it, this is our L circuit is equal to its denoted by town, and it's equal to the induction CE divided by the total resistance. We'll call that arson teak. This is going to be equal to the induction CE, which we just found. And the total resistance is the resistance in the copper wire are subsea, which we found back in part a here 1.3 owns plus are so by which is tthe e internal resistance. Um, here, which we're told is the, uh, internal resistance of the battery. Okay. And if you go back to page one, our supply is equal to 0.35 homes. So plugging all those values into this expression, we find that the time cost a is equal to 0.46 milliseconds, which is 0.46 times 10 to the minus three seconds Weaken box that in as the solution for E. Okay, starting a new page here for part s part, F asked us to find the maximum current that could be produced. We'll call that I necks. This is equal to the, uh, potential supplied by the battery, which is six volts divided by the total internal resistance. Excuse me. The total resistance, which of course again is equal to the resistance in the copper wire, plus the internal resistance playing those values into this expression, we find that the maximum current is equal to 3.6 in the units here. Hurry. Appears. Okay. And then for part G, it says, how long would it take to reach 99% of its maximum current or excuse me, 99.9% of its maximum current. So to do that, we can use the expression I the current at any given time is equal to ay Max. And this is for our l circuits, by the way, my next multiplied by one minus e to the minus t over town, which is talent being our time. Constant. Okay, so now we want to solve for, uh, for tea. So we're going to divide both sides by I, Max. Okay. And then start isolating the teeth. The time value. So this is e to the minus. T overtown is equal to one minus. I over I, max, but we're told that I is equal a 99.9% of I max. So this is going to be one minus 10.999 Since I over I, Max is going to be 0.999 So this is now, um, taking the natural log of both sides. We find that tea time is equal to minus the time constant towel times the natural log of one minus 0.999 So carrying out this expression, we find that the time here is equal to 3.2 milliseconds. We can go ahead and box set in a czar solution for part G. Now for Part H. It's asked us to find the energy stored in the induct er well, the energy stored in the induct er It will start a new page, says H. We'll call it use and l is equal to 1/2 times induct in CE claims the current I square. Okay, so, for I hear we're going to use my max for induct in CE it's, ah, the value that we found for Part D, which is 0.76 Miller Henry's and then for I will use. I'm actually going to say I is equal to my max. Okay, So, plugging all those values into this expression, we find that the potential here is equal to 4.9 mil it Jules, with just 4.9 times 10 to the minus three jewels. But we had induct Ins's and Miller Henry's so we can keep this and Millet Jules box it in is their solution for part H."}

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