Travelling at a speed of 2.0 10^5 m s ⁻¹ , a proton passes undeflected through crossed electric and magnetic fields that are perpendicular to each other. The proton's velocity is perpendicular to both fields. After the electric field is switched off, the proton travels in a circle of radius 4.0 cm . Determine the magnitudes of the electric and magnetic fields. Assume the mass of the proton is 1.6 10⁻²⁷ kg .

Step-by-step solution

When the proton moves undeflected in crossed electric and magnetic fields, the electric force balances the magnetic force: qE = qvB E = vB When the electric field is switched off, the proton moves in a circular path under the influence of the magnetic field. The radius of the circular path is given by: r = mv qB Rearranging for the magnetic field B : B = mv qr Substituting the given values ( m = 1.6 10⁻²⁷ kg , v = 2.0 10^5 m s ⁻¹ , q = 1.6 10⁻¹⁹ C , r = 4.0 10⁻² m ): B = 1.6 10⁻²⁷ 2.0 10^5 1.6 10⁻¹⁹ 4.0 10⁻² B = 2.