# Simple magnetic fields problem

1. Feb 24, 2013

### lovelyrwwr

One component of a magnetic field has a magnitude of 0.0266 T and points along the +x axis, while the other component has a magnitude of 0.0648 T and points along the -y axis. A particle carrying a charge of +2.96 × 10-5 C is moving along the +z axis at a speed of 4.71 × 103 m/s. (a) Find the magnitude of the net magnetic force that acts on the particle. (b) Determine the angle that the net force makes with respect to the +x axis.

I got the correct answer for part A, which is simply 9.77x10-3 N.
Fx ~ 0.003708 N
Fy ~ 0.009034 N
F = sqrt(Fx^2 + Fy^2)
F = 9.77x10-3 N

However, I am not getting the correct answer for B.

I thought it was simply:
theta = InverseTangent(Fy/Fx) = 67.68 deg but this wasn't correct.

Then I thought I had to add 90 to get theta = 67.68 + 90 = 157.68 deg since F is perpendicular to both field (B) and velocity (V), but this was not correct either. What am I doing incorrectly? Thank you!

Last edited: Feb 24, 2013
2. Feb 25, 2013

### Basic_Physics

According to the vector product (hope you know this version) the force on the positive charge
is perpendicular to both the velocity vector and the magnetic field vector:

$\vec{F}$ = q $\vec{v}$ $\times$ $\vec{B}$

, that is you turn the velocity vector towards the magnetic field vector. The force vector points in the direction in which a righthand screw advances due to this turning. This means $\vec{F}$ will be perpendicular to the plane containing the two vectors $\vec{v}$ and $\vec{B}$. Or you may know the right-hand rule for determining the direction of $\vec{F}$.

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