In this question, we're told we have a strip of silver, it is not point not five thick and 20 wide. It is in a magnetic field of 9.8 Kessler's and there are 5.85 times 10 to the 28 free electrons per cubic meter. There is a current of 10 amps through the strip. So what we want to find first is the drift velocity of the electrons. In order to do this, we need to use the expression I the current is equal yeah. To A. And a V. D. Which is the drift velocity, is the charge of the electron and is the number of them present and A is the area of the strip. So we know all of these values. We want to find the drift velocity. So what we need to do is rearrange his expression to the drift velocity is equal to the current over A. N. E. We've been given all of these in the question, we know that A. is equal to the thickness times the width. So we can just substitute this into here. And what we can say is that as a result A is equal To one times 10 to the -6 meters squared. And we can use that in this expression here as T. W. So now we've got this, we can substitute into the original expression for V. D. We know the current is 10 Amps has given in the question and we divide this theory by this area. We just calculated equal to T. W. So the thickness times the width. So 1.6, one times 10 to the -6. Rather Multiplied by N 5.85 times 10 to the 28 as given in the question, Yeah. Multiplied by the charge on an electron. So 1.6 times 10 to the -19. Yeah. Okay, This gives us an answer for the drift velocity equal to 1.07 Times 10 to the -3 m/s. Yeah. In the next part of the question, we want to calculate what the hall voltage measured by the meter is. We know that the expression for the hall voltage Is equal to V eight equals B. The magnetic field strength multiplied by the current over the thickness times a number of charge carriers, times a charge on the electron. So, plugging in the values we've got we know again the B field the magnetic field is equal to not .8. The current is equal to 10. The thickness is equal to not point not five. And the number of child carriers is 5.85 times 10-28. And E is again 1.6 times 10 to the -19. So we can sub in these values. So not .8 for B. Not applied by 10 for the current over, Not point not five times 10 to the -3 as it's needing to convert into meters. Multiplied by M. 5.85 times 10 to the 28. Yeah, Multiplied by the charge 1.6 times 10 to the -19. Which gives us a value equal to 17.1 μv as the whole voltage. Yeah, the final part of this question wants us to calculate which side of the volt meter would be at the highest potential. So we know that the Lawrence force is given by F equals Q V cross B. In the case that these vectors the velocity is V and B is the magnetic field. Again, we know that as a result of the right hand factor product rule. so crossing these two vectors, the force on the electron will be towards the left, and as a result, the direction of the hall current will be towards the right, leaving the left side of the volt meter to have a higher potential reading. Therefore, the left of the volt meter will be at the highest potential. Mhm. Yeah. Mhm. Yeah.