Question

A mass spectrometer is designed to separate protein fragments.The fragments are ionized by removing a single electron and then enter a 0.80 T uniform magnetic field at a speed of 2.3

Answer 1

Here's the COMPLETE answer:

Once the protein fragment is ionized (i.e. loses ONE electron) its electric charge is:

q = charge = 1.602 x 10^-19 Coulombs

The magnetic force exerted on the protein ion when it enters the mass spectrometer is given by:

F(magnetic) = q * v * B

Where

q = charge = 1.602 x 10^-19 C
v = protein ion's velocity = 2.3 x 10^5 m/s
B = strength of the magnetic field = 0.80 T

The situation simplifies because the direction of the magnetic field and velocity are perpendicular.

But we also note that the magnetic force is balanced by the centripetal force of the particle traveling IN A CIRCLE:

F(centripetal) = m*v^2 / r

m = mass of protein ion = 85 * mass of proton = 142 x 10^-27 kg
v = velocity = 2.3 x 10^5 m/s
r = radius of ion's path

When we equate the two forces we get:

m*v^2 / r = q*v*B

or

r = (m*v) / (q*B)

But the distance from the entrance of the ion to the point where it exits the magnetic field is 2*r (the DIAMETER of the circle is makes). Therefore:

D = distance = 2*r = (2*m*v) / (q*B)

= (2 * 142*10^-27kg * 2.3*10^5m/s) / (1.602x10^-19C * 0.80T)

so

D = 0.51m = 51 cm

So your answer is that the particle will exit the magnetic field 51 cm from where it entered.

P.S. I'm assuming the protein is only ionized ONCE - i.e. only ONE electron is stripped from the protein. This is usually the case.