A 150-g ball containing 4.00x10^8 excess electrons is dropped ...

Question

A 150-g ball containing 4.00x10^8 excess electrons is dropped into a 125-m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.

Accepted Answer

Hey everyone. So this problem is working with magnetic fields. Let's see what they are giving us and what they're asking from us, we have a brass sphere of a given mass. It has a given number of excess electrons. The sphere is released without initial speed. It's another way of saying it's released from rest from the top of the tower at 100 m below the release point. The sphere encounters a magnetic field of a given strength and that magnetic field is directed towards the east. We are asked to determine the strength and direction of the force exerted on the sphere by the magnetic field, we're told to neglect air resistance. So the first thing you can do here is recall that the force from a magnetic field is given by F equals the absolute value of Q V B sign data. So let's look at each of these variables one by one. So Q is our charge. We can recall that Q equals N E where N is the number of excess electrons and E is our the charge of an electron or electric constant. So we actually have everything we need to solve that the number of electrons is three times 10 to the seven. And E We can recall is a constant negative 1.6 times 10 to the -19 columns. And when we plug that in, We get negative 4.8 Times 10 to the -12 columns. Alright, so that's Q VR velocity will come back to be is the magnitude of our magnetic field that's given to us in the problem as 1.18 tesla and then sine of theta R theta is 90 degrees. We know that because the magnetic field is directed towards the east and the sphere is dropped in a vertical direction. So between the vertical and the horizontal, the angle is 90 and we can recall that the sign of degrees is one. And so this term goes to one. So we have everything we need to solve for the force except for V. We can recall now through the conservation of energy that delta K plus delta U equals zero, where delta K is the change in our kinetic energy. And delta U is the change in potential energy. If we recall that K equals one half MV squared and you equals M G H, we actually now have everything we need to solve for V for velocity to plug back into our first equation for force. So delta K and delta U, the delta is always going to be the final minus the initial. So we're going to rewrite that as K F minus K I plus UF minus UI equals zero. Let's take our potential energy over to the other side, K F minus K I equals subtract that both of those terms to the other side and simplify the negatives. And that's going to be you I minus U F now. And so we'll plug in our equations for kinetic and potential energy. And so that looks like one half M V F squared minus one half M V I squared equals M G H one minus M G H I M G H F. Okay. So we can simplify this out before we start plugging in any numbers. We know that our initial velocity is zero because it was released from rest. That entire term goes to zero. We have one half M V F squared equals M and G or constants so we can pull those out and then we'll have a chat I minus H F and then the max is also canceled. So for H I minus H F just go off to the side here and recall that we fell 100 m from the problem, right? So H I were going to call zero and H F is going to be 100 m down from that. So that's minus 100 m. So H I minus H F is going to be zero minus 100 m or positive 100 m. And so now we have everything we need to solve for our final velocity. So V F is going to equal the square root of two times gravity. Constant recall is 9.8 m per second squared Times HI - HF which we have just proven is m. And so when we plug that in To our calculator, we get 44. m/s. So we're going to go back up to equation one. So this was the absolute value of QVB and plug in everything that we have solved for there. So we Know that Q is negative 4.8 times 10 to the -12 columns. The absolute value of that is just going to be 4.8 Times to the -12 columns. two QV We just solve for is 44.3 m/s. And then be the magnitude of our magnetic field is 1.18 Tesla. We plug all of that into our calculator and we get 2. times 10 to the minus 10 news. So that is the magnitude of the force. Let's go back up to our answers. And we can see that A and B are incorrect, so we can eliminate those. And now we need to think about the direction of the force. So from our right hand rule, we know that the charges out of the page because it's negative, our magnetic field is directed to the east. So when we put our pointer point your finger up and our magnetic field to the east, our thumb points to the north. So that is the direction of our force. So the correct answer for this problem is C that's all we have for this one, we'll see you in the next video.

Key Concepts

Magnetic Force on a Moving Charge

Charge of Electrons

Kinematics of Free Fall

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