- #1
fruitbubbles
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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