Acceleration of a Particle Due to Magnetic Field

  • #1

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 (2i + 3j + 4k) mT. What is the acceleration of the particle in component form?

Homework Equations

F = q(vXB)
F = ma

The Attempt at a Solution

First convert all values to SI units:
mass = 0.002 kg
q = -6x10^-6 C
v = 3000 i m/s
B = (0.002i + 0.003j + 0.004k) T

Find vXB:
vXB = -12j + 9k

Plug vXB and q into F = q(vXB) and find a:

F = (-6x10^-6)(-12j + 9k) = ma
a = (7.2x10^-5 j - 5.4X10^-5 k) / 0.002
a = 0.036j - 0.027k m/s^2

However the correct answer provided to me is:
(36j - 27k) m/s^2

I think i made a mistake converting the units for B from mT to T. I'm not sure how to convert units for vector components.

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  • #2
Never-mind I found my mistake. I didn't convert the mass correctly.

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