How Do You Calculate Magnetic Force in Three Dimensions?

AI Thread Summary
To calculate the magnetic force on an electron moving in the X-Y plane within a three-dimensional magnetic field, start by expressing the velocity and magnetic field in component form. The magnetic force is determined using the formula F = q(v x B), where q is the charge of the electron, v is its velocity, and B is the magnetic field. Each component of the force can be calculated separately: the x component is 2.4 * 10^-14 N, the y component is -8 * 10^-15 N, and the z component is 4 * 10^-15 N. The resultant magnetic force acting on the electron is (2.4 * 10^-14 N, -8 * 10^-15 N, 4 * 10^-15 N). Understanding this approach clarifies how to handle magnetic forces in three dimensions.
EvaBugs
Messages
19
Reaction score
0
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?
 
Physics news on Phys.org
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!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top