Question

A singly ionized (one electron removed) 40K atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 km/s to pass through undeflected when the magnetic field is 2.50×10^-2 T. The ions next enter a second uniform magnetic field (B') oriented at right angles to their velocity. 40K contains 19.0 protons and 21.0 neutrons and has a mass of 6.64×10^-26 kg.
(a) What is the magnitude of the electric field in the velocity selector?
E = _______ V/m

(b) What must be the magnitude of so that the ions will be bent into a semicircle of radius 12.5 ?
B' = ________ T

Answer 1

Answer:

(a) 112.5 V/m

(b) 0.01494 T

Explanation:

velocity, v = 4.5 km/s = 4500 m/s

magnetic field, B = 2.5 x 10^-2 T

mass, m = 6.64 x 10^-26 kg

charge, q = 1.6 x 10^-19 C = 1.6 x 10^-19 C

(a) The force due to the magnetic field is balanced by the force due to the electric field

q E = B q v

E = B v = 2.5 x 10^-2 x 4500 = 112.5 V/m

(b) radius, r = 12.5 cm

$r = \frac{mv}{B'q}$

$B' = \frac{mv}{qr}$

$B' = \frac{6.64\times10^{-26}\times4500}{1.6\times 10^{-19}\times 0.125}$

B' = 0.01494 T