Finding acceleration of a particle due to the magnetic field

AI Thread Summary
A particle with a mass of 2.0 mg and a charge of -6.0 µC moves in a magnetic field while traveling along the x-axis at 3.0 km/s. The force on the particle is calculated using the equation F = qv x B, leading to two components of force in the y and z directions. The initial calculations yield an acceleration of -0.027 m/s² in the y-direction and -0.036 m/s² in the z-direction, but these results do not match the correct answer of (36j - 27k) m/s². The discussion emphasizes the importance of correctly applying the cross product and considering the angles between the vectors involved. Clarification on vector operations and the right-hand rule is sought to resolve the discrepancies in the calculations.
fruitbubbles
Messages
19
Reaction score
0

Homework Statement


A particle (mass 2.0 mg, charge –6.0 µC) moves in the positive direction along the x-axis with a velocity of 3.0 km/s. It enters a magnetic field of (2.0i + 3.0j + 4.0k) mT. What is the acceleration of the particle?

Homework Equations


F = ma to find the acceleration
F = qv x B to find the force of the magnetic field

The Attempt at a Solution



So I figured that since the velocity is in the positive x direction, and the x component of the magnetic field is also in a positive direction, the angle between them is 0 and so I can just not worry about the x component.

To get Fj, I have Fj = qvBsin(90), since j is in the positive y direction and the velocity is in the positive x direction. This gives me a force of -5.4 x 10-5, which if I divide by the mass, I get a = -.027 m/s2. I do the same for the k component, which I think the angle separating the field and the velocity is also 90 degrees, so Fk = qvBsin(90), which I get to be -7.2 x 10-5, and if I divide by the mass, I get a = -0.036 m/s2.

so my final answer would be a = (-.027j - 0.036k)m/s2. That is not one of my answer choices, which are as follows:

However, the correct answer is (36j - 27k)m/s2. My order of magnitude is wrong but so are the vectors (my answer has 27j, the correct is 27k, and the same for 36). I'm really bad with dealing with vectors so I'm sure that's the problem, but can anyone guide me to the correct answer?
Thank you
 
Physics news on Phys.org
fruitbubbles said:
So I figured that since the velocity is in the positive x direction, and the x component of the magnetic field is also in a positive direction, the angle between them is 0 and so I can just not worry about the x component.
... why not check that by computing the determinant for the cross product?
 
Simon Bridge said:
... why not check that by computing the determinant for the cross product?
..I do not have the slightest idea as to what that is.
 
fruitbubbles said:
..I do not have the slightest idea as to what that is.
Okay, I just looked at my notes and see what you mean, but I don't understand why sometimes I can just do the cross product of A x B as ABsin(theta) and sometimes I have to do all of that. I have no clue what it means
 
For two vectors given by:
##\vec r_1 = a\hat \imath + b\hat\jmath + c\hat k##
##\vec r_2 = d\hat \imath + e\hat\jmath + f\hat k##

Then the cross product is given by:
$$\vec r_1\times\vec r_2 = \begin{vmatrix} \hat\imath & \hat\jmath & \hat k\\ a & b & c \\ d & e & f \end{vmatrix}$$
 
  • Like
Likes fruitbubbles
If you correctly guess the angle, you can use ##ABsin\theta## to get the magitude and the right-hand rule to get the direction.

Go through the working for the determinant a step at a time and compare with your assumptions from before. i.e. can you safely ignore the x component of the magnetic field? Can you take the other components separately like you did? Did you absently switch two components over?
 
Last edited:
  • Like
Likes fruitbubbles
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...
Back
Top