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

a) 351351.35m/s

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

Step-by-step 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=(E)/(B)=(1.95*10^(5)V/m)/(0.555T)=351351.35(m)/(s)`

b) the mass-charge ratio is given by:

`(m)/(q)=(rB)/(v)=((6.61*10^(-3)m)(0.555T))/(351351.35m/s)=1.044*10^(-8)(kg)/(C)`

hope this helps!!