# Magnetic forces

1. Mar 24, 2014

### vanitymdl

A particle with mass 1.81×10−3kg and a charge of 1.22×10−8C has, at a given instant, a velocity v =(3.00×104m/s)j^. produced by a uniform magnetic field B =(1.63T)i^+(0.980T)j^

Find the value of the expression qvB/m (the magnitude of a when v is perpendicular to B ), where q is the magnitude of the charge, v is the magnitude of the velocity, B is the magnitude of the magnetic field, and m is the mass of the particle.

qvB/m =_____________m/s2

Attempt to problem:

F = m*a = q*(v cross B)

a = q/m * (v cross B)

v cross B = (0i + 30000j m/s) cross (1.63i + 0.980j T) = 0 i + 0 j - 48900 k m/s*T

a = 1.22 * 10^-8 C / 1.81 * 10^-3 kg * (0i + 0j - 48900 k m/s*T) = 0i + 0j - 0.3296 k m/s^2

But it keeps saying that the problem does not depend on I,j,k

2. Mar 25, 2014

### paisiello2

The question is asking for the magnitude.

3. Mar 25, 2014

### rude man

The problem asks for a when v is perpendicular to B. But you have computed a at the moment when v = 3e4 j. v is then not perpendicular to B since v dot B ≠ 0.

EDITed last part of original post.

Last edited: Mar 25, 2014