Electromagnetism - right hand rule

[avsj]

Homework Statement

A proton traveling at a speed of 3.0 x 10^6 m/s travels through a magnetic field of strength 3.0 x 10^-3 T, making an angle of 45 degrees with the magnetic lines of force. What force acts on the proton?

Homework Equations

F = QvB

The Attempt at a Solution

I arrived at F = 1.44 x 10^-15N by simply plugging in. Then using the 45 degrees, I assume the force I found is the hypotenuse so I solve for an adjecnt using trig to get the correct answer of 1.0 x 10^-15 N but I don't understand this conceptually. Why is the force at 45 degrees weaker, and why am I solving it this way?

Thanks a lot

 

[meopemuk]

Homework Statement

A proton traveling at a speed of 3.0 x 10^6 m/s travels through a magnetic field of strength 3.0 x 10^-3 T, making an angle of 45 degrees with the magnetic lines of force. What force acts on the proton?

Homework Equations

F = QvB

The Attempt at a Solution

I arrived at F = 1.44 x 10^-15N by simply plugging in. Then using the 45 degrees, I assume the force I found is the hypotenuse so I solve for an adjecnt using trig to get the correct answer of 1.0 x 10^-15 N but I don't understand this conceptually. Why is the force at 45 degrees weaker, and why am I solving it this way?

Thanks a lot

The correct Lorentz force equation is

\mathbf{F} = Q \mathbf{v} \times \mathbf{B}

where \mathbf{v} \times \mathbf{B} stands for the cross-product of two vectors. Since you know the angle \alpha = 45 degrees between \mathbf{v} and \mathbf{B}, you can use standard formula for the length of the cross-product vector

| \mathbf{v} \times \mathbf{B}| = |\mathbf{v}| |\mathbf{B}| \sin \alpha

to obtain the magnitude of the force |\mathbf{F}|.

Eugene.