An electron moves in the magnetic field B⃗ = 0.600î T with a...

An electron moves in the magnetic field \(\vec{B} = 0.600\hat{i}\text{ T}\) with a speed of \(1.50 \times 10^7\text{ m/s}\) in the directions shown in the figure. For each, what is magnetic force \(\vec{F}\) on the electron? (Figure 1) For help with math skills, you may want to review: The Vector Cross Product Express vector \(\vec{F}\) in newtons in the form of \(F_x, F_y, F_z\), where the x, y, and z components are separated by commas. Express vector \(\vec{F}\) in newtons in the form of \(F_x, F_y, F_z\), where the x, y, and z components are separated by commas.

Transcript

00:01 In this problem we have an electron so the charge is minus 1 .6 10 to the power minus 19 cool arms that moves in a magnetic field 0 .590 tesla in the exit direction or 0 .590 one zero zero tesla with a speed 1 .00 10 to the power 7 meters per second in two different configurations and for each of these configurations we are going to find the force on electron in the cartesian form okay the first configuration is this we have x here y here z here and the velocity is like this in the negative y direction.

01:16 So we have v equal to 1 .00 10 to the power 7.

01:24 The direction is negative y head so it is 0 minus 1 .0 in unit of meters per second.

01:33 Now let us compute the force using the cross -prolux.

01:37 So force is equal to q times v cross b.

01:46 Okay, we have q times this cross product is the determinant of the following matrix.

01:53 We have x, y, head, z head.

01:57 The components of v are 0 minus 1, 10 to the power 7, 0, and the component.

02:08 Components of b are 0 .590 0 .0.

02:18 Okay, this cross - this determinant will give us the following vector.

02:24 So the x component is we remove this column and this row and compute the determinant of the rest.

02:33 For y component we cover this column and this row, we have the negative determination.

02:40 Of the rest and for z component we cover this column and row and compute the determinant of this 2x2 matrix this will give us the following vector we have 0 0 0 5 .9 10 to the power 6 and if we now multiply it by the electron charge each component we are going to obtain 0 0 minus 9 .44 10 to the power minus 10 in units of newton's or minus 9 .44 10 to the power minus 10 z head neutons so this is the force acting on our electron in the first configuration in the second case we have the following.

03:50 X here, y here, z here, and the velocity vector is defined on the yz plane, making this angle 45 degrees with the negative y axis.

04:10 So we have v equal to 1 .0010 to the power 7...

Question

Please give Ace some feedback