At a point high in the Earth’s atmosphere, He²⁺ ions in a conc...

Question

At a point high in the Earth’s atmosphere, He²⁺ ions in a concentration of 2.8 x 10¹²/m³ are moving due north at a speed of 2.0 x 10⁶ m/s. Also, a 7.0 x 10¹¹ / m³ concentration of O⁻₂ ions is moving due south at a speed of 6.2 x 10⁶ m/s. Determine the magnitude and direction of the current density $\vec{j}$ at this point.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem beneath the equator C A two plus ions with a concentration of 3.0 multiplied by 10 to the power of 12 ions per cubic meter, move east at 2.2 multiplied by 10 to the power of 6 m per second. Meanwhile fe three plus ions with a concentration of 6.0 multiplied by 10 to the power of 11 ions per cubic meter, move west at 5.5 multiplied by 10 to the power of 6 m per second, determine the current density vector J. So it appears the final answer that we're ultimately trying to solve for is we're trying to determine what the current density is. So that means that we are solving for the numerical value of vector J. So now that we know that we're trying to solve for the numerical value for vector J, we're trying to figure out what the current density is for this particular problem. Let's read off our multiple choice sensors to see what our final answer might be noting that all of our current density values are in units of amps per meter squared. And we're also given a direction. So A is 0.53 westward, B is 0.53. Eastward, C is 0.85. Westward D is 0.85 eastward. Awesome. So first off, let us write down all of our known variables. So for the C A two plus ions, we know what we're given as an like what variables were given. For this particular part of our problem, we're told that the concentration N is equal to 3.0 multiplied by 10 to the power of 12 ions and it's ions per cubic meter. Awesome. We're also told that the charge which we're gonna call Q is equal to two because there's two ions, it, since it's two plus. So we need to do two, multiplied by 1.60 multiplied by 10 to the power of negative 19 and its units are Kums per ion. So that will mean that Q is equal to. Well, actually, for now we'll just keep it like that instead of multiplying it out. Awesome. And this, once again, we are multiplying by two since C A two plus has a charge of positive two E since there's two positive electrons. So, and we also know the velocity which we're gonna call VD is equal to 2.2 multiplied by 10 to the power of 6 m per second. And let us make a note in blue that it's moving eastward. So it's moving to the east. Awesome. So moving right along here. So now considering our fe three plus ions condition, we are told that the concentration for this case, which once again is N is equal to 6.0 multiplied by 10 to the power of 11 ions per meters cubed. We're also told that the charge Q in this case is equal to three because fe three plus has a charge of positive three E. So when you take three multiplied by 1.60 multiplied by 10 to the power of negative 19. And once again, its units will be coulombs per ion fantastic. And moving right along here, we also know the velocity which we're gonna call VD. And the velocity in this case is equal to 5.5 multiplied by 10 to the power of 6 m per second. In this case, it is moving to the West. Awesome. So now we need to note that we are trying to determine the magnitude and direction of the current density vector J at this point. So we need to now at this, due to that, we now need to recall and use the formula for the current density vector J. So let us recall that vector J is equal to N multiplied by E multiplied by VD. Awesome. So then we need to note and recall that for multiple types of ions, the total current density will be the sum of the individual current densities. So with that in mind, let us consider the eastward motion to be positive and the westward motion to be negative. So solving for our C A two plus ions first, so let's make a note of that, that we're dealing with their C two plus ions condition first, that we can go ahead and write that vector JC is equal to N multiplied by C multiplied by QC which so we need to be very clear here. So NC A is our concentration for our C A two plus condition multiplied by our charge for our C A two plus condition multiplied by V DC A which plugging in our known variables will give us that this is all equal to 3.0 multiplied by 10 to the power of 12 ions per cubic meter multiplied by two multiplied by 1.6 or 1.60 multiplied by 10 to the power of negative 1919 poems per ion and lastly multiplied by 2.2 multiplied by 10 to the power of 6 m per second. Awesome. So let me plug that into our calculator. We will find that vector JC A is equal to positive 2.112. So that's when we round to three decimal places, amps per meter squared. Awesome. So now we need to solve for the fe three plus condition. Awesome, which will give us when we plug this into our calculate or actually when we do the same exact methodology and then plug our expression. Once we plug in our variables into our calculator, we will find that vector JFE is equal to NFE multiplied by Q fe multiplied by VDFE is all equal to plugging known variables. 6.0 multiplied by 10 to the power of 11 ions per cubic meter multiplied by three, multiplied by 1.60 multiplied by 10 to the power of negative 19 columns per ion uh sub columns per annum. And then we have to multiply this by our velocity which in this case, since it's moving westward, we have to make sure we have our negative sign. So it's negative 5.5 multiplied by 10 to the power of six meters per second. Awesome. So that will mean that our final answer for vector JFE is equal to when we plug this into our calculator. And we round to three decimal places like our vector JC A answer, we will get negative 1.584. And once again, its units will also be amps per meter squared. So now we can officially solve for our final answer which is the net current density. So the net current density which is vector J is equal to vector JC plus vector JFE, which is equal to when we plug in our known variables, positive two, 2.112. So 2.112 amps per meter squared. And then we need to add this to negative one point so 1.584 and its units will be amps per meter squared. So when we plug that into our calculator, we will find that our final answer is equal to when we round to two decimal places. That will mean that our final answer rounded to two decimal places is approximately equal to 0.53 and its units will be amps per meter squared and its direction will be eastward. So I'll just write the word east for sure. And that's because our final answer is positive. So that means it has to be the eastward direction and that's it. That's our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter B, 0.53 amps per meter squared and it's in the eastward direction. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Current Density

Charge Carriers

Vector Addition

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