What Angle Gives a Magnetic Force of 1.8 Times the Original Force?

[kpangrace]

When a charged particle moves at an angle of 19° with respect to a magnetic field, it experiences a magnetic force of magnitude F. At what angle (less than 90°) with respect to this field will this particle, moving at the same speed, experience a magnetic force of magnitude 1.8F?



can someone please tell me why doing a simple 19/f=x/1.8f cross-multiply and divide won't work?


it's my last problem on this dang homework!

 

[Tide]

Why should it work? You have no basis for writing that proportion.

You need use the Lorentz force equation for charged particles moving in a magnetic field. Does this look familiar: [itex]\vec F = q \vec v \times \vec B[/itex]? Given [itex]\vec v[/itex] and [itex]\vec B[/itex] could you determine the magnitude of [itex]\vec F[/itex]? How does it relate to the angle between the vectors?

 

[quicknote]

I understand your frustration with this problem. However, I would like to point out that using the equation 19/F = x/1.8F may not work because it assumes that the magnitude of the magnetic force is directly proportional to the angle of the charged particle's movement. In reality, the magnitude of the magnetic force is affected by other factors such as the strength of the magnetic field and the velocity of the charged particle. Therefore, a more accurate approach to solving this problem would be to use the formula F = qvBsinθ, where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and θ is the angle between the particle's velocity and the magnetic field. By rearranging this formula, you can determine the value of θ that would result in a magnetic force of 1.8F. I hope this helps and good luck with your homework!