1. Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field Physicists designed the mass spectrometer to assist the chemistry folks in determining the mass of a given particle (atoms, isotopes, molecules, etc) This is a simplified version of the inner workings of a mass spec machine Ionizing Source - - + + E B

2. Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field You can see 3 main areas that make up the mass spectrometer A. the ionizing source, B. a chamber with crossed electric and magnetic fields, and C. a chamber containing only a magnetic field Ionizing Source - - + + E B A B C

3. Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field Ionizing Source - - + + E B Electric field, E is upwards (negative plate on top, positive plate on the bottom) Magnetic field, B is out of the page These fields are “crossed” or perpendicular to one another. Crossed Electric & Magnetic Fields 1 st opening 2 nd opening Note: The 1 st and 2 nd openings are directly opposite one another. If a particle was shot out of the ionizing source through the 1 st opening it would travel straight through the 2 nd opening as well.

4. Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field Consider how a similar electric field alone would act on a negatively charged particle Now consider how a magnetic field alone would act on the same particle - - + + v FeFe v FmFm Combine these two fields in the mass spectrometer’s first chamber and the particle feels equal but opposite forces FmFm FeFe vRvR vRvR vRvR

5. Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field This is precisely what happens… The ionizing source charges the sample particles using either heat or electric current and ejects them through that 1 st small opening into the first chamber The electric & magnetic fields are quickly adjusted so that the opposing forces acting on the charged particles cancel each other out and keep them moving in a straight line at a constant speed right into the next chamber. Ionizing Source - - + + E B Negatively charged particle

6. We can use the effect of the crossed electric & magnetic fields in this chamber to determine the actual speed of the charged particles If E = F e /q, then we can The magnetic field creates rearrange to solve for F e an equal & opposite force F e = E qF m = q v B F e = F m E q = q v B ** q’s cancel Speed of the particle through v = E/ B crossed electric & magnetic fields Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field So, combining the two equations

7. Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field Once the charged particles pass through the 2 nd opening, they enter and area where they are subjected to a magnetic field alone. The magnetic force acting on the particles are no longer cancelled out.. The magnetic force acts on the moving particles causing them to spin until they hit the inner wall or leave the field entirely. LHR#4 determines the direction of the force & path the particle will travel Ionizing Source - - + + E B Negatively charged particle

8. The idea that magnetic fields cause charged particles to travel in circular paths suggests centripetal motion is involved!!! The magnetic force, F m, must be equal to the centripetal force, F c. F m = q v B F c = m v 2 r F m = F c q v B = m v 2 ** v ’ s cancel r Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field So, combining the two equations This equality can be rearranged to solve for either radius of curvature, r, or mass of the particle, m, depending on your interest.

9. If we are the physicist designing the device, we’d be interested in calibrating the machine using known masses and checking their radius of curvature, r. q v B = m v 2 ** v ’ s cancel r Radius of Curvature, r r = m v qB Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field Where v = velocity (m/s) m = mass (kg) q = charge of the particle (C) B = magnetic field strength (T) r = radius of curvature (m)

10. If we are the chemist, using the mass spectrometer to determine and unknown mass, we’d rearrange to isolate m. q v B = m v 2 ** v ’ s cancel r Mass of the particle, m m = qBr Mass Spectrometer: A Practical Use for Moving Charges in a Magnetic Field Where = velocity (m/s) m = mass (kg) q = charge of the particle (C) B = magnetic field strength (T) r = radius of curvature (m) v v

11. Sample Problem 1 A positively charged particle is emitted from the ionization chamber through crossed electric and magnetic fields of 375 V/m and 4.85x10 3 T respectively. If it experience a radius of curvature of 2.96x10 -2 m and carries a charge of 6.4x10 -19 C, what must be the mass of the particle?

12. Sample Problem 1 A positively charged particle is emitted from the ionization chamber through crossed electric and magnetic fields of 375 V/m and 4.85x10 3 T respectively. If it experience a radius of curvature of 2.96x10 -2 m and carries a charge of 6.4x10 -19 C, what must be the mass of the particle? Givens: E = 375 V/m Solve for m B = 4.85x10 3 Tm = qBr q = 6.4x10 -19 C r = 2.96x10 -2 m v

13. Sample Problem 1 Givens: E = 375 V/m Solve for m B = 4.85x10 3 Tm = qBr q = 6.4x10 -19 C r = 2.96x10 -2 m First we need to find velocity & then mass = E = 375 V/m = 7.73x10 4 m/s m = qBr B 4.85x10 -3 T =(6.4x10 -19 C)(4.85x10 3 T)(2.96x10 -2 m) 7.73x10 4 m/s = 1.19x10 -21 kg So the mass of the unknown particle is 1.19x10 -21 kg. v v v

14. Sample Problem 2 What would be the radius of curvature for a 4.51x10 -26 kg particle that has lost 3 electrons when is passes out of crossed electric and magnetic fields of 1275 V/m and 2.68x10 -5 T respectively into the magnetic field alone?

15. Sample Problem 2 What would be the radius of curvature for a 4.51x10 -26 kg particle that has lost 3 electrons when is passes out of crossed electric and magnetic fields of 1275 V/m and 2.68x10 -5 T respectively into the magnetic field alone? Givens: E = 1275 V/m Solve for r B = 2.68x10 -5 T r = m q = lost 3 e - = 4.8x10 -19 C qB m = 4.51x10 -26 kg v

16. Sample Problem 2 Givens: E = 1275 V/m Solve for r B = 2.68x10 -5 T r = m q = lost 3 e - = 4.8x10 -19 C qB m = 4.51x10 -26 kg v First we need to find velocity & then radius = E = 1275 V/m = 4.76 m/s r = m B 268T qB =(4.51x10 -26 kg)(4.76m/s) (4.8x10 -19 C)(2.68x10 -5 T) = 1.67x10 -2 m So the radius of curvature for this particle is 0.0167 m. vv