How Do You Calculate Magnetic Force in Three Dimensions?

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SUMMARY

The discussion focuses on calculating the magnetic force acting on an electron moving in the X-Y plane with a velocity of (5 * 10^5 m/s, 30 degrees) in a uniform magnetic field represented by B = (6, -2, 1). The formula used is F = qv x B, where q is the charge of the electron (1.6 * 10^-19 C). The resultant magnetic force is calculated by determining the x, y, and z components separately, resulting in a total force of (2.4 * 10^-14 N, -8 * 10^-15 N, 4 * 10^-15 N).

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  • Understanding of vector mathematics and cross products
  • Knowledge of electromagnetic theory, specifically Lorentz force
  • Familiarity with the concept of magnetic fields and their components
  • Basic understanding of trigonometric functions, particularly sine
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Hello.

I am currently working on this problem:
"An electron moves in the X-Y plane with a velocity v = (5 *10^5 m/s , 30 degrees). The uniform magnetic field i the region is B= ( 6 i - 2 j + 1 k ). Find the resultant magnetic force on the electron in component form."

I know that F = qv x B = qvB sin (teta)

What mixes me up is the fact that B has 3 components. How should I begin?
 
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You could begin by writing v in terms of it's 3 components. Then do the cross product.
 


Hello there,

Thank you for reaching out for help with this problem. It seems like you have a good understanding of the formula for calculating the magnetic force on a moving charged particle. You are correct in using the formula F = qv x B = qvB sin (teta).

In this case, the magnetic field B has three components, which can be written as (6, -2, 1). In order to find the resultant magnetic force on the electron, we can break down the formula into its component form. This means we will calculate the x, y, and z components of the force separately and then add them together to find the total force.

Starting with the x component, we have qvB sin (teta) = (1.6 * 10^-19 C)(5 * 10^5 m/s)(6 T)(sin 30 degrees) = 2.4 * 10^-14 N. This is the force acting in the x direction.

Similarly, for the y component, we have qvB sin (teta) = (1.6 * 10^-19 C)(5 * 10^5 m/s)(-2 T)(sin 30 degrees) = -8 * 10^-15 N. This is the force acting in the y direction.

Lastly, for the z component, we have qvB sin (teta) = (1.6 * 10^-19 C)(5 * 10^5 m/s)(1 T)(sin 30 degrees) = 4 * 10^-15 N. This is the force acting in the z direction.

To find the total force, we simply add these components together, giving us a resultant magnetic force of (2.4 * 10^-14 N, -8 * 10^-15 N, 4 * 10^-15 N). This is the force acting on the electron in the X-Y plane with a velocity of (5 * 10^5 m/s, 30 degrees).

I hope this helps you understand how to approach this problem. Good luck with your calculations!
 

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