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An electron moving the direction shown below by the Vector The magnitude of the clectron" velocity is 4X 10' nVs The magnitude and direction of the magnet...

Question

An electron moving the direction shown below by the Vector The magnitude of the clectron" velocity is 4X 10' nVs The magnitude and direction of the magnetic ficld are also shown in the diagran The anele between and B is 33 degrees: Determine the magnitude and direction of the force acting on the clec33 degreesB=2T

An electron moving the direction shown below by the Vector The magnitude of the clectron" velocity is 4X 10' nVs The magnitude and direction of the magnetic ficld are also shown in the diagran The anele between and B is 33 degrees: Determine the magnitude and direction of the force acting on the clec 33 degrees B=2T



Answers

An electron moves with a velocity of $1.0 \times 107 \mathrm{~m} / \mathrm{s}^{1.0} \times 10^{7} \mathrm{~m} / \mathrm{s}$ in the $x-y$ plane at an angle of $45^{\circ}$ to both the $+x$ and $+y$ axes. There is a magnetic field of $3.0 \mathrm{~T}$ in the $+y$ direction. Calculate the magnetic force (magnitude and direction) on the electron. SSM Example $19-1$

Hi in the given problem there is a uniform magnetic field directed upward in the plane of paper. But it is like this. Now in this magnetic field we are taking Electrons at three places moving In three different directions. This is the location A. At which the electron is moving just towards right the magnetic release upward. Then another location B. At which electron's velocity is making an angle of 20 degree with the horizontal. And at the third location see the lost city of electron is making an angle off artie degree from the horizontal. So this velocity vector will be making an angle of the extreme degree with the direction of magnetic mm. The magnitude of magnetic failure is 0.40. Yes, lot speed of electron That is given as 8.0 into 10 days. To the par five m for a second. Now using the basic expression for the force experienced by a charged particle in our magnetic field. And that is actually magnetic Lawrence force. They're using the expression for magnetic Lawrence force which says F is equal to you into We cross the Oh we can see this is a physical to Q. We be I'm peter were theater will be the angle between velocity vector and magnetic field vector. And for electron this force in vector form it will be given us minus E. Until we cross B. To find the direction of this force. First of all, we will use right hand screen rule over this cross product of V with B. And whatever the direction we will get. We will make reverse of it due to the negative sign. Now, in the first part of the problem is the electron which has been put at location a angle between Magnetic vector and velocity vector here in this location at this location is 90°. So magnitude of that force will be given by Or charge over the electron. 1.6 in two 10 dish par minus 19 colon. The velocity this is eight into 10, part five Nita per second for magnetic field 0.40 Tesla and signed 90. Which will come out with just one. The magnitude of this force comes out to be 5.12 Into 10 dish the par -14. Newton answer for the first part of this problem and to find its direction, direction of we crossed the and that is using right and screw rule. This direction is outward perpendicular to the plane of paper. So, direction of force experienced by electron due to the negative sign of electron, it will be reversed. So it will be into the plane of paper. Now, in the second part of the problem, it is clear from the figure that angle between velocity vector and magnetic field vector here, This angle will be 90 -20 means 70 degrees. So magnitude of the force E B. Sign 70 degrees for every job it will be same as they have obtained In the first part of the problem. 5.12 Into 10 for -14. Newton. Then for sign, 70 degree, it will be 0.774. So finally, this force experienced by the electron in the second position comes out to be we point 96 Into a 10 for -14. Newton. one of the answer for the second part of the problem and other direction. Using right and screw rule. We cross B. Direction of Crosby is into the plane of paper. So, due to negative sign over the electron, F will be outward perpendicular to the plane of paper. Finally, in the third location at that location and girl Between the magnitude of vector and velocity vector here, this is 60°. So As Tita is 60° the magnitude of force E. V. B signed 60 degree for every B. Again, 5.12 Into 10. This bar -14 Newton And for science 60° is ruled three x 2. So finally, this force experienced by the electron at location C. This is 4.43 Into 10 days par -14 new 10. one of the answer for the third part of this problem. And finally, to find the direction of force as we cross B is again outward here using rock and screw rule. So, F will be into the plane of paper again due to the negative sign of electron. Thank you

This question covered the concept of magnetic force on a charged particle moving in a magnetic field. The magnetic force of B on a charge particles. The magnitude of charge into the cross productive the velocity reactor with magnetic field. The magnetic force on the electron is, the charge on electron is minus 1.67 to 10 days minus 19 colon. And diversity of the electron is 1.0 into 10 days, five m/s along X. direction that would escape. And to the magnetic field is to Tesla along. Does that exist therefore other magnetic force on the electronic -3.2 and 2 10 raise -14. Cool um meters per second. An ex cap project cap is minus white cap. Okay. Therefore the magnetic force is 3.2 into 10 days -14. Why can't therefore the magnitude of the force on the electron is 3.2 into temporaries -14 Newton. Okay, this is the newton newton. And the direction is a long positive. Why access? And the correct option is option. The

Here for the solution. This is a free body diagrams. This is the beef electrons, the we and they are the access X axis y axis and your taxes and to solve the next step. Yeah, we know the equation that the force only the tone is equal to f electrical two QB multiply by B here. We know the question that force on electron affected equal to Q. We multiply by B and here we substitute all the values in the equation. Uh, the value of Q is 1.6 multiplied by 10 24 minus 19. The value V is 8.75 Multiply by 10 to the power bye I kept. And for the B s 0.45 Jacob. And by solving this, we get difficult to six 0.3. Sorry. 6.3. Multiply by 10 to the power minus 14. K 6.3 multiplied by 10 to report minus 14 Cake Cab Newton. So this is a force of on electrons. Yeah, The direction of the force is towards minus along with the jet access here in this direction of force. Well, it's too weird. Minus alone do their access So this is the solution? Yeah, this is a magnet. If it be access, are that X and Y And here for the solution force on the electron. We know that electrical two q v multiply by B and held by securing the value of Cuba that is 1.6 multiplied by 10 to depart minus 19. It is V 8.75 multiplied by 10 to the power of five icap multiply by page at 0.4 five's a cap and by solving this week, it 6.3 multiplied by 10. To department is 14 k cap and the direction of forces towards minus along the third axis. Please go through this thing.

In this problem. We have a electron moving at a constant velocity and were asked to find the magnetic field at four different points a through D on to find the Cartesian coordinate system with ex horizontal. Why barnacle and see out of the page were given at a velocity of 10% of the speed of light to three significant figures as well as each position would like to find us two microns away from the electron. So to find this, we use the abuse of our law for a constant a particle with constant velocity which is be of our it's equal to charge of the moving particle times the permitted ity constant, uh, multiplied by the magnitude of the velocity over four pi times, gift magnitude of the distance squared and this is all multiplied by the cross product between the direction of the velocity in the direction where we're trying to find the magnetic field. Now we know for every point in this problem, this whole quantity is a constant. We can call this be not for the velocity. It is for each point going to be pointing in the white hat direction and the distance to the point we're trying to find the magnetic field. We can write this as co sign data are ex hat direction plus sign. What's data are in the Why had direction? Now we know that. Why Hat Cross? Why Hat is going to be zero So we can on Lee the only concern ourselves with the X component of this position vector to find the magnetic field. And if we plug in the values for our at point A we get be of our a is equal to do you not times co sign 60 degrees times Why hat cross except and this evaluates to negative Z hat and co sign of 60 degrees is 1/2 so we can write this as negative be happy be knocked over to in the Z at direction And since our B has the same X component as our A, we know that this is also the same as the magnetic field at RB. Now for our C, The position on Lee has an ex Capone component so we can write this as negative. Be not see him. And for our d this on. Lee has a Y hat component and so we can also right that be r D is zero. So now we have to plug in all of the given values for the distance, the velocity, the permeability constant, a swell as the charge of the electron. With this, we get that the constant be not is equal to negative 1.20 times 10 to the minus seven Tesla on This is negative because the charge of the electron is negative and notice when we plug this into all of these values for B r, A, b, R B and B R C, it cancels with this negative sign and we get that all of these fields point in the positive Z at direction out of the screen on. We wrote this as with three significant figures because that is all the significant figures were given for the velocity in the distance.


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