Question

A proton goes undeflected in a crossed electric and magnetic field (the fields are perpendicular to each other) at a speed of 2.0 x 105m/. The velocity is perpendicular to both the fields. When the electric field is switched **, the proton moves along a circle ol radius 4.0cm. Find the magnitudes of the electric and the magnetic fields. Take the mass of the proton = 1.6 x 10-27kg.

Answer 1

MP = 1.6 × 10–27Kg

v = 2 × 105m/* r = 4cm = 4 × 10–2 m

Since the proton is undeflected in the combined magnetic and electric field. Hence force due to both the fields must be same.

i.e. qE = qvB => E = vB

Won, when the electricfield is stopped, then if forms a circle due to force of magnetic field