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An 18 gauge copper wire (diameter 02 mm) carries current with current density of 2.10 x 106 A/m The density of free electrons for copper is 8.5 x 1028 electrons per...

Question

An 18 gauge copper wire (diameter 02 mm) carries current with current density of 2.10 x 106 A/m The density of free electrons for copper is 8.5 x 1028 electrons per cubic meter:Calculate the current in the wire_Express your answer with the appropriate units_ValueUnitsSubmitRequest AnswerPart BCalculate the magnitude of the drift velocity of electrons in the wire_ Express your answer with the appropriate units_ValueUnitsSubmitRequest Answer

An 18 gauge copper wire (diameter 02 mm) carries current with current density of 2.10 x 106 A/m The density of free electrons for copper is 8.5 x 1028 electrons per cubic meter: Calculate the current in the wire_ Express your answer with the appropriate units_ Value Units Submit Request Answer Part B Calculate the magnitude of the drift velocity of electrons in the wire_ Express your answer with the appropriate units_ Value Units Submit Request Answer



Answers

An 18-gauge copper wire (diameter 1.02 mm) carries a current with a current density of $3.20 \times 10{^6} A/m{^2}$. The density of free electrons for copper is $8.5 \times 10^{28}$ electrons per cubic meter. Calculate (a) the current in the wire and (b) the drift velocity of electrons in the wire.

Okay, so we know the current density was just J here. Can we go to currently about my area? So therefore current, I should be good j A. So current density in this case is given as 1.5 times 10 to par six MPR per meter square and the area should be equal to pi times the diameter divide by two to the power to and I'm during this case is a good one point. Also millimeter if we call into meter is he with the one point 0 to 10 to the power of 93 meter, so eventually have errors. You go to a 2.6 year old one pie test into a power opening. Seven meters were so now it's fucking better The equation to determine occurrence which is equal to 1.50 times 10 to the power of six MPR for a meter, score multiple by 2.601 pie times 10 to the power of Negus seven meters square and this will give us the occurrence is equal to one point 23 m Pierre. So not it's determine the drivel last the so we know current committee Goto on q b d a is the electron density, which is evil. The one I'm sorry, a point by 10 to 10 to the power between the A premier, a cube in his question. And one of the current is he with 1.23 m. Pierre And we know que Here is the charge of the chart carriers, which is the charge of electrons, Things Charles Caris veterans and another charge of the Lateran is 1.6 times 10 to the power of neck. 19 columns in the area is you go to a 190.601 attends 10 to the power of the next seven years, for it does not arrange me here with the drivel drivel. Last DVD is a good guy over on the way, So now that's plugging better. The equation. To determine the drift velocity, which is equal to 1.23 m. Pierre, divided by a point by tens 10 to the power of 28 per meter cubed and times 1.6 has tend to the power of the neg knighting Coolum and Mobile by 2.601 high times 10 to the power of the next seven meter square. And then he moved. Uh, I'm here to the middle. Also look better. Okay, so, uh, this will give us the dreadful. Last year is equal to one point 11 times. Tend to the power off. Negative for meter for a second. We can worry too. Millimeter per second is equal to 1.1 war times 10 to the power off connective. One millimeter per second, which is equal to zero point. We don't want one millimeter for a second, and these are the answers for this question.

High in the given problem. Current density is given as J is equal toe 3.0 into 10 days to the power six, um, Pierre per meter square and then city off electrons. This wire he's given us and is equal toe 8.5 into 10. They should part 28 electrons per meter, que diameter off the wire is 1.2 millimeter, so it's radius will be R is equal toe half off the diameter means 1.2 by two means this is zero point 51 millimeter or we can say this is 0.51 into 10. Dish part minus 3 m. No. In the first part of the problem, we have to find the current passing through this wire. So using the basic expression for current density that is current passing per unit area off the conductor. We get an expression for current as the product off current density will do area. Or we can say this is Jay in two pi R square, so plugging in the non values for current density. This is 3.20 into 10. This part six um Pierre, 4 m square multiplied by area, which is 3.14 into 0.51 into 10. Dish par, minus three meter to the whole square. So finally, this current passing through the conductor comes out Toby two point 61 ampere. And here it becomes toe answer for the first part off the problem. No. In the second part of the problem, we have to find drift velocity of the electrons in that conductor. So again, using the expression for current density j physical toe current passing through unitary of prosection. And the relation for current in terms off drift velocity is I is equal. Tow N E a v e d. Where n is the number density of electrons is the charge of our one electron A is area and we d is the drift velocity. So, using that expression for the philosophy, consult Toby G. Current density divided by n into e. So again plugging in or non values For Jay, this is 3.20 into 10 rich part six m Pierre per meter square for end. This is 8.5 into 10 ish for 28 into for electronic chart. This is 1.6 into 10 days, part minus 19 to learn. So finally, this drift velocity comes out. Toby 2.35 into 10. Dish par, minus 4 m per second. And here it becomes the answer for the second part off the problem. Thank you.

Number four in the first part. We know that current density J is equal to Corin upon a DEA. So current densities, given its one born fight into and to the power six. This is equal to Karen, upon idea by by four in 20 point 001 they go to It's quiet. Solving this, we will get the current I equals two 1.25 and we are now in the second part of the question, their velocities to be completed. We know that current density is equal to end times you in tow their philosophy. The current densities 1.5 indoor and to the power six and the value off N for Cooper is and 815 into 10 to the power 28. You have really 1.6 in tow, 10 to the power minus 19. Fulham's in Judah below. Stevie Solving this, we will get the velocity of equal to 0.11 millimetre files again

Hi number density off free electrons in the Given Conductor is 8.5 into tentage part 28 electrons per neater. Que electric field applied across its terminals is 0.0 600 Newton per column at the temperature off 20.0 degrees Celsius. There is no use off temperature just to maken idea about the resistive ity off the copper at this much temperature and this resistive ity at this much temperature can be found. Toby 1.72 into 10. Dish par minus eight on 2 m. No. In the first part of the problem, we have to find drift velocity of the electrons, for which we use an expression for the current density. In terms off electric field that has given us J is equal to the ratio of electric field applied to the resistive ity off the conductor and another expression is jade. In sequel to current density means current passing through unique area. Now the relation for the current in terms off your philosophy is I is ableto n e a we d There n is the number density of electrons e charge over one electron and we desert the philosophy here area will be canceled out so J is equal to N E V. D. So using that, using these two equations, we conclude that e by raw is equal tow N E into VD. Hence we get an expression for the philosophy, which has given us e by N E into role, so plugging in all known values. Reedy here comes out to be 0.6 for electric field applied for number density of electrons. This is 8.5 into 10. December 28 into Elektronik Charges 1.6 into 10. Rich for minus 19 into resistive ity. We know at 20 degrees Celsius this is 1.72 into 10. Dish par minus Kate. So finally this VD comes out Toby 2.56 in tow. 10 dish for minus 4 m per second. Or we can say we did Visible tau 0.256 milli meter per second. And here this is the answer for the first part off the problem. No. In the second part of the problem, we have to find potential difference applied across the ends off this conductor. If the length of the conductor is 20.0 centimeter, or we can say this is 0.20 m. So using the relation between electric field and electric potential, which sales electric field is given as potential Grady End means we buy health. So using that we is equal toe be in tow. Hence this we comes out Toby the product off 0.6 world perimeter or Newton per colon in 20.20 m. So finally, this potential difference across the answer the conductor Yes, found to be 0.12 wall. And this is the answer for the second part off the problem. Thank you.


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