At a particular instant, charge q1 = +4.80 ×10^-6 C is at the point (0, 0.250 m, 0) and has velo

At a particular instant, charge q1 = +4.80 ×10^-6 C is at...

At a particular instant, charge $q_1 = +$4.80 $\times$ 10$^{-6}$ C is at the point (0, 0.250 m, 0) and has velocity $\vec{v_1}$ = (9.20 $\times$ 10$^5$ m/s)$\hat{\imath}$. Charge $q_2 = -$2.90 $\times$ 10$^{-6}$ C is at the point (0.150 m, 0, 0) and has velocity $\vec{v_2} =$ (-5.30 $\times$ 10$^5$ m/s)$\hat{\jmath}$. At this instant, what are the magnitude and direction of the magnetic force that $q_1$ exerts on $q_2$?

Transcript

00:01 Okay, we're looking at question 2856, where we have two point charges, q1 and q2, that are moving at different velocities.

00:10 So q1 is located at 0 .0 .25 meters, 0.

00:16 So that's just right up on the y axis here.

00:19 And it has velocity 9 .2 times 10 to 5 meters per second in the i -hat direction.

00:24 And the i -hat direction is the positive x direction.

00:27 So that's going to be pointing straight to the right like this.

00:30 Now charge 2 has a charge of minus 2 .9 times 10 to minus 6 coulums.

00:42 And this is columns.

00:45 And it's moving with velocity minus 5 .3 times 10 to the 5 meters per second.

00:52 So that's in the j hat direction.

00:55 So j hat direction is the y direction.

00:57 So since it's negative, that's going to be going in the minus y.

01:00 Like this and it's at point 1 .150 zero so that means it lies at 0 .15 meters on the x -axis and the other two coordinates are zero.

01:11 So we're asked at this point what are the magnitude and direction of the magnetic force that q1 exerts on q2 so to solve this we're going to need our equation that the force is equal to q v cross b and when we think about this we want the force that's exerted on q2 from the magnetic field from q1.

01:36 So this magnetic field is gonna be the magnetic field coming from q1.

01:41 So here we want q2 and v2 to be what we're considering here.

01:50 Now this magnetic field, this b will be a b coming from q1.

01:54 So here's the expression for the magnetic field from a moving point charge.

02:00 So here again, we're talking about q1.

02:04 So we're going to need to put in q1 here.

02:07 And again, v1 here, because we're talking about this point charge, creating a magnetic field that q2 is going to experience.

02:15 So at this point, q2 is a distance r from q1.

02:21 So i have listed that over here.

02:25 So our first step is going to be to figure out what r is.

02:29 So we have a nice little triangle here with the x coordinate is 0 .15.

02:35 The y is 0 .25.

02:37 So we can use the pythagorean theorem where we know that r squared is just going to be equal to 0 .25 squared plus 0 .15 squared.

02:51 And when we plug that in and solve, we can get that r is going to be 0 .29.

03:01 So we have our r now.

03:04 Now, next what we need to do is think about this cross product.

03:10 So we have our r here for our cubed, but we need to know this v1 cross r.

03:17 So we know the direction of v1 and we know the direction of r, but, and we know the magnitude of r.

03:24 But the cross product, these two vectors aren't going to be perpendicular.

03:27 So the cross product is going to be equal to v1 times r times the sign of this angle here.

03:35 So if we label this angle, let's call it phi.

03:43 So this is this angle phi right here.

03:45 Now our expression for the magnetic field can be written here in terms of v1 times r times the sign of this angle phi.

04:01 So now we, and before we actually get into this, let's just see quickly that there's a factor of r up here and a factor of r down here.

04:09 So i'm just gonna quickly cancel out one of these factors and turn this into r squared, just while we have it here.

04:15 So now we need to know what sign of phi is.

04:19 Well, sign of phi, if you look at this, this angle, we could make another triangle, by just drawing a line right over here if we wanted...

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