contestada

A proton is moving at 2.1 x 10 m/s clockwise through a constant and perpendicular magnetic field. The radius of curvature through the field is 0.6 meters. What is the strength of the magnetic field?

Respuesta :

Explanation:

For this question the magnetic force provides the force required for the circular motion

Equating

Magnetic force = centripetal force

Bqvsinx = mv^2/r

as magnetic field is perpendicular x=90

Bqvsin90 = mv^2/r

Bq = mv/r

B= mv/rq

Then replace r= 0.6 , v= 2.1ms and mp from formula sheet and you can obtain B which is magnetic field intensity

Otras preguntas