Question

A proton with a speed of 2.7 × 106 m/s is shot into a region between two plates that are separated by a distance of 0.16 m. as the drawing shows, a magnetic field exists between the plates, and it is perpendicular to the velocity of the proton. what must be the magnitude of the magnetic field, so the proton just misses colliding with the opposite plate?

Answer 1

Hello Dear

in this case

electron will move in a circular path so centripetal force acting on it will work as magnetic force or simply

mv²/r = qvB

mv/r = qB

B = mvq/r

B = 1.67 x 10⁻²⁷ x 2.7 x 10⁶ x 1.6 x 10⁻¹⁹ /0.16

B = 7.21 x 10 ⁻⁴⁰ /0.16

B = 0.256 x 10⁻⁴⁰ = 2.56 x 10⁻⁴¹ T