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QUESTION 1Q1: circuit, on Average 2S0U0 clectrons pass through point along wlrc in [ms]; What is the magnitude ol this current?QUESTION ?02: L the cunent through- ...

Question

QUESTION 1Q1: circuit, on Average 2S0U0 clectrons pass through point along wlrc in [ms]; What is the magnitude ol this current?QUESTION ?02: L the cunent through- wire is 4 [mA] , what is the aVerage arount o tira takcn by elecuon pas $ through a point aong the Wire !QUESTION ]Which of the below palit Eive currents dhe uamc +- sign? A-An clecuon moving Itom lelt to ught proton mnoving Iromn Icft t0 right proton mOving trom rght t0 Iclt D - An elcctron moving (rom rielt Ia Ielt

QUESTION 1 Q1: circuit, on Average 2S0U0 clectrons pass through point along wlrc in [ms]; What is the magnitude ol this current? QUESTION ? 02: L the cunent through- wire is 4 [mA] , what is the aVerage arount o tira takcn by elecuon pas $ through a point aong the Wire ! QUESTION ] Which of the below palit Eive currents dhe uamc +- sign? A-An clecuon moving Itom lelt to ught proton mnoving Iromn Icft t0 right proton mOving trom rght t0 Iclt D - An elcctron moving (rom rielt Ia Ielt



Answers

$.$ ILW Figure 29-51 shows a snapshot of a proton moving at velocity $\vec{v}=(-200 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}$ toward a long straight wire with current $i=350 \mathrm{~mA} .$ At the instant shown, the proton's distance from the wire is $d=2.89 \mathrm{~cm} .$ In unitvector notation, what is the magnetic force on the proton due to the current?

Our system is mad up off two wires they are joined together. Joined together to calculate the electric potential difference between the two points. You first calculate their resistance. The potential difference between a point 0.1 and two is Delta. We want to their physical toe i r c where are sees the resistance off. Why are seeing similarly the potential difference between potential difference between the two and three points His i r d where these there are resistance off Weir d They're cross morning rates off energy dissipation will be P 12 musical toe I square R C and the p 23 physical toe I square rd What ails the problem? Ah, we write our see resistance ofwar seasonable to roll See time's LCD Bye bye. I r C square That is a two times 10 to the power minus six times one divided by by time 0.50 meter square. This uses 2.55 homes. Then we could find a Delta V 12 is equal to i r c. That is to em piers times 2.55 homes Excuse us 5.1 walls but be off the problem. Similarly, we confined it Resistance off Hawaii, which will be our D, is equal to roadie times l d Divided by bi rd square This is one times 10 to the power minus six times one meter. Divided by pi times 0.25 meters square This will you, us our de resistance off 5.9 homes from here. Then we can find the potential difference between Delta v 23 points going by I r D. That is a two MPR times 5.9 homes that isn't called thio talk 10.22 walls which is approximately tin walls but see off the problem. The power dissipated between the points one and two is given by P one to that is I square times r c thirties a 10 wolves body off the problem. The power dissipated between a bob 0.213 is Batou fees I square rd that is 20 walls

So this problem we have some junction and we're told that wire one will go into the junction with a current of went for camps. And then we have wire to coming out of the junction with the current I to as six five Yepes and for part A, we want to know the number of electrons flowing through some wire three, which we don't know. We're wired three is that But we do know that when we're dealing with currents flowing in through a junction, the current flowing into the junction I n has to equal the current coming out of the junction will call I out. So we know that the current in is 0.4 and the current out this 0.65 So to find the wire three. Since we know that both sides of the equation have to be equal, then we just need to subtract I out from i n. So we get 0.65 minus four, and this will give us the current and wire three, which will call I three. So we get 0.25 amps. So that's the current. And now we need to know the number of electrons passing through that wire. Well, we know that current point for wire 3.25 amps. Amps is really cool homes for second. So the amount of charge flowing through the wire per second. So if you just multiply by one over the charge off the electrons, we get the number of electrons and the charge of an electron is one point 602 times 10, the minus 19 columns of charge and, of course, charges canceled, and we're left with one. We multiply at 1.5 six times 10 to the 18th number of electrons for a second. Mhm, that's part. And now we need to know the direction of this wire three. That's for part B. So again, if you think about it, we have a junction we know going into already is I won. And coming out of it is I to and again going in and going out have to equal each other. I am. We already know this 0.4, and then coming out is 0.65 So to get these equations or these numbers to match, I three has to add two. I won, So we have four plus 25. Then we get them too much. So again, if you draw the junction, I won going in I to going out so that I three has to go in the same direction as I one, which is into the junction.

So in this problem. Uh, so we have the wire and we know that to the diamond. The damage of the wire is ah, 2.5. Maybe there's and pull. The lens of the wire is 14 meters and ah, we also know that the resistance is that point on forums. Ari is the point. Ah, all four homes. So in part, I want to find out the resistive ity of this wire. We know that the resistance, the relation between the resistance and the resistive it is ah, are equal row L over air. Aye. Ayes the cross section and a has expression pai d over to squid. So in this equation, the only on the verbal is this resistive ity, and you can see that the row equal, uh, 3.65 times 10 to the negative eighth o meter and in part B. Ah, So suppose supposed Ah, electoral magnetic. Supposed to like the electric field magnitude inside this wire is one point 28 lot meter. So want to find out the current inside this wire, So basically, we're because we're gonna have this electric feud. So the total potential difference on this wire is Ah, 1.28 times 14 meters. Right. So this is the total potential difference. And we use the potential difference. Divide by the resist The resistance are so this gives us the current I and the vet who is 1 72 and peers a Parsi. Want to find out the drift velocity off this Ah ah of the electrons inside the wire. So the expression of the drift velocity is the you equal I do. I buy end and el Rey eyes the current, which is the one we found here and is Ah ah! The density of the electron which is giving the problem is the 8.5 times 10 to the 28th. So is 8.5. I understand to 28th and ah, How is the lens off this? Ah, I'm sorry. Not l This is I'm sorry. Yeah, this is the okay And ease the charge of the electron which is 1.6 times 10 to and igniting squirms. So you can see that finally you equal 2.58 times 10 to the negative third meters a second

Our question says that the current density in a wire is uniform. It has a magnitude J as 2.0 temps into the six amperes per meter squared the wire like Ellis five meters and the density of the conduct of electrons in is a 50.49 10% of the 28 meters. Cute. How long does it take the electron on average to drift the length of the wire then. So first we're going to Let's do this under the line, we're going to need to calculate the, uh, drift velocity. So drift velocity is equal to the magnitude of J divided by the density and times the charge e plug those values in this expression and it comes out to 1.47 times 10 to the minus four meters per second. Okay, With that calculated, we can calculate the time it would take with something traveling at that velocity to travel the distance. L This is going to be l divided by the velocity v sub d clicking those values in this comes out to be 3.4 times 10 to the fore units. Here are seconds weaken box set in as our salute


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