An alpha particle moves undeflected through a region containing crossed uniform electric and magnetic fields. When the electric field is switched off, the alpha particle traces a circular path of radius 2 ~cm . If the magnitude of the electric field is 10^4 ~V/m , the magnitude of the magnetic field is _______ mT . Take the mass of the alpha particle as 6.4 10⁻²⁷ ~kg and its charge as 3.2 10⁻¹⁹ ~C .

Step-by-step solution

For the undeflected path of the alpha particle in crossed electric and magnetic fields, the electric force balances the magnetic force: qE = qvB v = E B When the electric field is switched off, the particle moves in a circular path under the influence of the magnetic field alone. The radius of this circular path is given by: r = mv qB Substituting v = E B into the radius equation: r = m qB ( E B ) = mE qB^2 Rearranging for B^2 : B^2 = mE qr Substituting the given values: B^2 = 6.4 10⁻²⁷ 10^4 3.2 10⁻¹⁹ 2 10⁻² B^2 =