How Fast Must You Drive to Experience a 210N Magnetic Force on Your Head?

[kiwikahuna]

Homework Statement

Oops, I forgot to add in the second part of the problem. Here's the original question plus the second part of the question.

The strength of Earth's magnetic field at the equator is approximately equal to 5e-5 Tesla. The force on a charge q moving in a direction perpendicular to a magnetic field is given by F = qvB, where v is the speed of the particle. The direction of the force is given by the right hand rule. Suppose you rub a balloon in your hair and your head acquires a static charge of 3 e-9 C. If you are driving east, how fast would you have to drive in order for the magnetic force on your head to equal 210 N (enough to knock you over)?

The Attempt at a Solution

Here's what I did but it seems too simple. Could anyone tell me if I'm actually doing this the right way or not? Thanks.
If force (F) = 210N
Then the formula for the velocity is
v = F / (Bq)
v = 210 N/ (5e-5 * 3e-9)
v = 1.4e15 m/s

 

[mezarashi]

Assuming that the magnetic field is pointing straight up, your solution appears "correct" minus one little detail, that is, the speed of light is 3e8 m/s, and knowing that it is impossible to bypass this limit, the answer would be impossible.

 

[m2003]

Your solution is correct! However, it is important to note that the speed calculated here is not physically possible for a human to achieve. This calculation is simply a theoretical value based on the given information. In reality, the force on your head would likely be much lower due to factors such as air resistance and the limitations of human movement. Additionally, the direction of the force would depend on the direction of your motion relative to the Earth's magnetic field, so the right hand rule would need to be applied to determine the correct direction of the force. Overall, your solution is a good demonstration of the relationship between magnetic force and velocity, but it is important to consider practical limitations in real-world scenarios.