How to Calculate Forces on an Electron at the Equator?

AI Thread Summary
To calculate the forces on an electron at the equator, the gravitational force can be simplified to F = mg, using g ≈ 9.81 m/s² instead of the full gravitational equation. For the electric force, the correct formula is F = qE, where E is the electric field strength of 100 N/C. The magnetic force is calculated using F = qv x B, with v being the electron's velocity and B the magnetic field strength of 50.0 µT. The right-hand rule is applicable for determining the direction of the magnetic force. Understanding these formulas and their applications is essential for accurately calculating the forces acting on the electron.
Pepsi24chevy
Messages
65
Reaction score
0
Here reads the problem. the equator, near the surface of the Earth, the magnetic field is approximately 50.0 µT northward, and the electric field is about 100 N/C downward in fair weather. Find the gravitational, electric, and magnetic forces on an electron in this environment, assuming the electron has an instantaneous velocity of 7.70 x10^6 m/s directed to the east.


Now for gravitational do i set it up by: (G * m1 * m2) / (d2)? If so what would i use for the distance.

For electric force i assume i set it up by: kq/r^2 ?

And then for magnetic forces i assume i set it up by: qvxB?

Also for the directions, would i use teh right hand rule for each one?

Any suggests towards the right way of doing this would be appreciated. Thanks
 
Physics news on Phys.org
For the gravitational force, if you were to use F=\frac{Gm_1m_2}{r^2}, then r would be the radius of the earth. However, this is unnecesary since around the surface of the Earth the terms \frac{Gm_1}{r^2} are nearly constant and given by g \approx 9.81m/s^2, so the force is just mg.

For the electric force, \vec{F}=q\vec{E} would be more appropriate.

Fort the magnetic force you have it right. The right hand rule is the way to find the direction of the magnetic force.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top