Understanding how mass spectroscopy works
Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields. 


Electric fields accelerate charged particles (linear acceleration). Magnetic fields deflect them (angular acceleration). 

Charged particles respond to electric fields and magnetic fields differently. Electric fields accelerate charged particles linearly in the direction of the flux. For example, if an electric field exists between two plates, one with a buildup of positive charge and the other with a buildup of negative charge, then positively charged particles will accelerate from the positive part toward the negative plate. The acceleration of the particle ($\vec{a}$) depends on the strength of the field ($\vec{E}$), the mass of the particle ($m$), and the charge on the particle ($q$). Since the acceleration is a scalar multiple of the electric field, they have the same direction.

$$\vec{F}=q\vec{E}$$
$$\vec{F}=m\vec{a}$$
$$q\vec{E}=m\vec{a}$$
$$\vec{a}=\frac{q}{m}\vec{E}$$

Magnetic fields are orthogonal to electric fields. Magnetic fields alter the angular acceleration of charged particles (i.e., they deflect them). Thus, charged particles have a circular trajectory in a magnetic field. The radius of the circular path ($r$) is dependent on the strength of the magnetic field ($\vec{B}$), the charge of the particle ($q$), the mass of the particle ($m$), and the velocity of particle ($\vec{v}$). The radius of curvature is proportion to the mass to charge ration ($m/q$).

$$|\vec{F}|=m\frac{|\vec{v}|^2}{r}$$
$$|\vec{F}|=|q\vec{v}||\vec{B}|$$
$$m\frac{|\vec{v}|^2}{r}=|q\vec{v}||\vec{B}|$$
$$r=\frac{m}{q} \frac{|\vec{v}|}{|\vec{B}|}$$

The acceleration of the particle depends on the velocity, magnetic field, radius, and mass. Since acceleration in a magnetic field is a scalar multiple of the cross product of velocity and field strength, the direction of acceleration is always changing.
$$\vec{F}=q(\vec{v}\times\vec{B})$$
$$m\vec{a}=q(\vec{v}\times\vec{B})$$
$$\vec{a}=\frac{q}{m}(\vec{v}\times\vec{B})$$


  Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.


Now that we know the relationship between radius of curvature and mass-to-charge ratio, this question is easy. The detector counts the number of particles that hit it. That is all that it can do. That is all that it needs to do. The hard work of separating the ions by m/q is done by the mass analyzer, which is the magnetic field in curved tube. The magnetic field scans (quickly) over a range of $\vec{B}$ to alter the curvature of the ions by their m/q. Since the tube has a fixed curvature, only the ions with the correct m/q will pass to the detector. All others will hit the walls of the tube since their radius of deflection would be too large or too small. Thus, since the instrument controls the strength of the magnetic field, and physics determines what m/q will have the correct deflection at that field strength, there is no need to calibrate the detector. 

More complex mass analyzers exist, for example the quadrupole mass analyzer. Instead of the typical bipolar magnetic field, it generates a quadrupolar magnetic field. Ions are thrown all over the place to hit a spherical detector. The physics is more complex, but the trajectory once again depends on m/q, so the detector only has to count hits in certain spots.