Posted by **Stryker** on Thursday, June 7, 2012 at 4:27pm.

An electron moves on a circular orbit in a uniform magnetic field. The strength of the field is 1.00ื10-5 tesla. The kinetic energy of the electron is 178.0 eV. (1 eV = 1 electron volt = 1.6 ื10−19 joules.) Calculate the radius of the orbit.

(electron parameters: the charge is -e where e = 1.602ื10-19 C; the mass is m = 9.11ื10-31 kg.)

- Physics -
**Elena**, Thursday, June 7, 2012 at 5:39pm
KE =mv²/2,

v= sqrt(2Ke/m) = sqrt(21781.610^-19/9.1110^-31) =7.910^6.

Lorentz force

F=qvBsinα,

Since the circular orbit, sinα = 1.

mv²/R = qvB.

R= mv/qB = mv/eB =

=9.1110^-317.910^6/1.610^-19110^5 =4.510^-10 m

## Answer this Question

## Related Questions

- physics - An electron has a kinetic energy of 4.2 x 10-17 J. It moves on a ...
- Vectors Math - The magnetic force (Vector FM)on a particle in a magnetic field ...
- Vectors Math - The magnetic force (Vector FM)on a particle in a magnetic field ...
- Vectors - The magnetic force (Vector FM)on a particle in a magnetic field is ...
- College physIcs - An electron travels at a speed of 4.0ื104 m/s through a ...
- Physics - A uniform electric field of magnitude 395 N/C pointing in the positive...
- Math - An electron accelerates from rest to the right, in a horizontally ...
- physics - An electron, moving with a speed, 6.8 km/s, enters a magnetic field ...
- physics - An electron of kinetic energy 41 keV moves in a circular orbit ...
- physics - speed of electron in magnetic field - A beam of electron is ...

More Related Questions