Magnitude of a magnetic field a a point

[Jstuff]

A point charge Q moves on the x-axis in the positive direction with a speed of 450 m/s. A point P is on the y-axis at y = +70 mm. The magnetic field produced at point P, as the charge moves through the origin, is equal to -0.8 μT k^. When the charge is at x = +40 mm, what is the magnitude of the magnetic field at point P? (μ0 = 4π × 10-7 T · m/A)

Hey guys so I am having trouble understanding this problem. Well I actually understand the problem, but I am having trouble applying the formulas.

I understand that to do that problem I need to use the formula for the magnetic field created by a moving point charge. B=(μ_0 q r x v)/(4π r^2) to solve for the charge.
I did this for position one where the degree between the angle will be sin(90)=1 and got -.00124C.
I then applied the equation again to solve for the magnetic field at the new position. My trouble I am having is the cross product between r and v at this locations. I put r into unit vector form and crossed it with v in the positive x direction, but I do not get the right answer.

 

[ehild]

A point charge Q moves on the x-axis in the positive direction with a speed of 450 m/s. A point P is on the y-axis at y = +70 mm. The magnetic field produced at point P, as the charge moves through the origin, is equal to -0.8 μT k^. When the charge is at x = +40 mm, what is the magnitude of the magnetic field at point P? (μ0 = 4π × 10-7 T · m/A)

Hey guys so I am having trouble understanding this problem. Well I actually understand the problem, but I am having trouble applying the formulas.

I understand that to do that problem I need to use the formula for the magnetic field created by a moving point charge. B=(μ_0 q r x v)/(4π r^2) to solve for the charge.

The formula for B is not correct. It should be $$\vec B = \frac {\mu_0}{4\pi} Q \frac{\vec v\times\hat r}{r^2}$$
so you have to cross the velocity with the unit vector pointing from the charge to P.
ehild

 

[Jstuff]

The formula for B is not correct. It should be $$\vec B = \frac {\mu_0}{4\pi} Q \frac{\vec v\times\hat r}{r^2}$$
so you have to cross the velocity with the unit vector pointing from the charge to P.
ehild

Okay, so then I am crossing (450i) x (.O4i-.07j)1/.08. Correct?

 

[ehild]

If 0.08 in the denominator stands for $\sqrt{65}$ then it is approximately correct.

 

[HalfLostAndSearching]

Hi there,

It looks like you are on the right track with using the formula for the magnetic field created by a moving point charge. However, there are a few things to keep in mind when applying this formula:

1. The charge Q in the formula represents the magnitude of the charge, not the position. So you will need to use the value of the charge given in the problem (in this case, it is not given but you can solve for it using the information given).

2. The vector r represents the distance from the moving charge to the point where you want to calculate the magnetic field. In this case, the point P is at (0,70 mm), so you will need to calculate the distance from the charge at (40 mm,0) to (0,70 mm).

3. The vector v represents the velocity of the moving charge. In this case, the charge is moving in the positive x-direction with a speed of 450 m/s, so the velocity vector would be (450 m/s, 0).

4. When taking the cross product between r and v, remember that the result will be a vector in the direction perpendicular to both r and v. In this case, the cross product should give you a vector in the positive z-direction, which is represented by the unit vector k^.

Once you have all these values, you can plug them into the formula and solve for the magnitude of the magnetic field at point P. I hope this helps and good luck with your problem!