Question

A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser speeds are deflected either to the left or to the right. After leaving the velocity selector, particles enter a purely magnetic field. A particle's radius of motion is then used to find the mass-to-charge ratio of the particle, which in turn can identify it. Taken altogether, this device is called a mass spectrometer. The illustrated electric field is directed to the right with magnitude 1.95 times 10^5 V/m and the magnetic field is directed into the screen with magnitude 0.555 T.

a. Determine the speed v of the undeflected charged particle.
b. After passing through the velocity selector, the charged particle moves in a circular path with a radius of r = 6.61 mm. Determine the particle's mass-to-charge ratio. m/q = kg/C

Answer 1

Answer:

a) 351351.35m/s

b) 1.044*10^{-8}kg/C

Explanation:

a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.

$F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}$

b) the mass-charge ratio is given by:

$\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}$

hope this helps!!