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

[tex] \mathbf{F} = Q \mathbf{v} \times \mathbf{B} [/tex]

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

[tex] | \mathbf{v} \times \mathbf{B}| = |\mathbf{v}| |\mathbf{B}| \sin \alpha [/tex]

to obtain the magnitude of the force [itex]|\mathbf{F}| [/itex].

Eugene.

 

[m2003]

I can explain the concept of the right hand rule and how it applies to this problem. The right hand rule is a tool used to determine the direction of the force acting on a charged particle moving through a magnetic field. It states that if you point your right thumb in the direction of the particle's velocity and your fingers in the direction of the magnetic field, the force acting on the particle will be in the direction pointed by your palm.

In this problem, the proton is moving at an angle of 45 degrees with the magnetic field. This means that the force acting on the proton will also be at a 45 degree angle with the magnetic field. Using the right hand rule, we can determine the direction of this force. If we point our thumb in the direction of the proton's velocity (towards the right), and our fingers in the direction of the magnetic field (upwards), our palm will be facing towards the left. This means that the force acting on the proton will be towards the left.

Now, to calculate the magnitude of this force, we use the equation F = QvB, where Q is the charge of the particle (in this case, a proton with a charge of +1.6 x 10^-19 C), v is the velocity of the particle (3.0 x 10^6 m/s), and B is the strength of the magnetic field (3.0 x 10^-3 T). Plugging in these values, we get a force of 1.44 x 10^-15 N, as you have calculated.

However, since the force is acting at an angle of 45 degrees, we need to use trigonometry to find the component of this force in the direction of the particle's motion. This is why you used the cosine function to find the adjacent side of the triangle. This gives us a force of 1.0 x 10^-15 N, which is the correct answer.

In summary, the right hand rule helps us determine the direction of the force acting on a charged particle in a magnetic field, and trigonometry helps us find the component of this force in the direction of the particle's motion. I hope this helps to clarify the concept for you.