Velocity Components and Magnetic Force

[Soaring Crane]

Homework Statement

A particle with charge 9.42×10−8 C is moving in a region where there is a uniform magnetic field with a magnitude of 0.440 T in the +x-direction. At a particular instant of time the velocity of the particle has components -v_x = 1.70×10^4 m/s, -v_y = 3.19×10^4 m/s, v_z = 5.83×10^4 m/s.

a. What is the y-component of the force on the particle at this time?

b. What is the z-component of the force on the particle at this time?

Homework Equations

F = q*v*B*sin theta

The Attempt at a Solution

In drawing the y and z components,

The y-component is pointing to the south, so the force’s y-component is pointing upward (positive?)?

The z-component is pointing upward out of the plane, so the force’s z-component is pointing north (positive?)?


As for calculating the individual force components, what else must be done aside from applying the F = q*v*B definition? (Just using F_y = v_y*B*q or F_z = v_z*B*q is incorrect, right?)

Thank you for any guidance.

 

[apelling]

For a full mathematical treatment of this type of problem it is necessary to construct a vector for velocity and magnetic flux density and find their vector cross product. But I don't think going into that would help you.

One thing to emphasise is that the x y and z directions are the same for forces, velocities and any other vector. Some of your statements suggest that you think that for example F_z and v_z are not in the same direction. They most definitely are. Maybe I have misunderstood you but it is an important point.

One way to tackle this is to imagine there are three identical charges of 9.42x10^-8 C. One is moving in the x direction at 1.7x10^4, one in the y direction at 3.19x10^4 and one in the z at 5.83x10^4. For each particle work out the force on it. The x moving particle will have no force since v and B are parallel. The y moving particle will have a force in the z direction by Fleming's LH motor rule and the z moving particle will have a y direction force again by FLHMR. Then you simply need to understand that these 3 forces( including the zero force) all apply to one particle moving with the original vector components.

(You very nearly had the right answer in the beginning)

 

[m2003]

Yes, you are correct in your understanding of the direction of the force components. The y-component of the force will be positive and pointing upward, while the z-component will be positive and pointing north.

To calculate the force components, you will need to use the equation F = q*v*B*sin theta, where theta is the angle between the velocity and magnetic field vectors. In this case, since the velocity is only in the x, y, and z directions and the magnetic field is only in the x-direction, the angle between them will be 90 degrees for both the y and z components.

So, for the y-component, the force will be F_y = q*v_y*B*sin 90 = q*v_y*B. Similarly, for the z-component, the force will be F_z = q*v_z*B*sin 90 = q*v_z*B.

Plugging in the given values, we get F_y = (9.42×10^-8 C)(3.19×10^4 m/s)(0.440 T) = 1.16×10^-2 N and F_z = (9.42×10^-8 C)(5.83×10^4 m/s)(0.440 T) = 2.44×10^-2 N.

So, the y-component of the force is 1.16×10^-2 N and the z-component is 2.44×10^-2 N.

I hope this helps clarify the solution for you. Remember to always pay attention to the direction of the force components and use the correct equation for calculating them. Keep up the good work!