a) The magnitude of the manetic field is solvable using theequation
\(\displaystyle{r}={m}\times\frac{{V}}{{{\left|{q}\right|}\times{B}}}\)
, where r is the radius of theion's motion, m is the ion's mass, V is the ion's velocity, q isthe ion's charge, and B is the magnetic field acting on theion.

\(\displaystyle{\frac{{{1}}}{{{2}}}}{M}{\left({V}{e}{l}{o}{c}{i}{t}{y}\right)}^{{{2}}}={q}{\left({V}{o}{<}{a}\ge\right)}\)</span>

PSK\frac{1}{2}\times (3.92\times 10^{-25}kg) \times V^{2} = (3.2 \times 10^{-19} C) \times 100kVZSK

\(\displaystyle{V}={4.04}\times{10}^{{{5}}}\frac{{m}}{{s}}\)

\(\displaystyle{r}={m}\times\frac{{V}}{{{\left|{q}\right|}\times{B}}}\)

\(\displaystyle{1}{m}={\frac{{{\left({3.92}\times{10}^{{-{25}}}{k}{g}\right)}\times{4.04}\times{10}^{{{5}}}\frac{{m}}{{s}}}}{{{\left({3.2}\times{10}^{{-{19}}}{C}\times{B}\right)}}}}\)

B = 2.02Tesla

b) Since current is measured in coulombs per second, the totalcharge moved per hour is the mass per hour over the mass per ion(which gives us the number of ions per hour), multiplied by thecharge per ion. This value is then divided by 3600seconds (perhour) to attain the current.

(massCollected / massPerIon) * CoulombPerIon / TimePeriod= Current

\(\displaystyle{\frac{{{\left\lbrace{.1}\times{10}^{{-{3}}}{k}{g}\right\rbrace}{\left\lbrace{3.92}\times{10}^{{-{25}}}{k}{g}\right\rbrace}}}{\times}}{\frac{{{\left\lbrace{3}\times{10}^{{-{19}}}{C}\right\rbrace}{\left\lbrace{3600}{s}\right\rbrace}}}{=}}{I}\)

I = .0227Amps

c) Energy dissipated in the cup is just a function of the kineticenergy dissipated for each ion times the number of ions collectedduring the period.

PSK\frac{1}{2}M(Velocity)^2 \times (massCollected / massPerIon)= EnergyDisspated \frac{1}{2} \times \frac{1}{2} \times (3.92\times 10^{-25}kg) \times (4.04\times 10^{5})^{2} \times \frac{.1\times 10^{-3}kg}{3.92\times 10^{-25}kg}) = EnergyZSK

Energy = \(\displaystyle{8.16}\times{10}^{{{9}}}{J}{o}\underline{{e}}{s}\)

\(\displaystyle{\frac{{{1}}}{{{2}}}}{M}{\left({V}{e}{l}{o}{c}{i}{t}{y}\right)}^{{{2}}}={q}{\left({V}{o}{<}{a}\ge\right)}\)</span>

PSK\frac{1}{2}\times (3.92\times 10^{-25}kg) \times V^{2} = (3.2 \times 10^{-19} C) \times 100kVZSK

\(\displaystyle{V}={4.04}\times{10}^{{{5}}}\frac{{m}}{{s}}\)

\(\displaystyle{r}={m}\times\frac{{V}}{{{\left|{q}\right|}\times{B}}}\)

\(\displaystyle{1}{m}={\frac{{{\left({3.92}\times{10}^{{-{25}}}{k}{g}\right)}\times{4.04}\times{10}^{{{5}}}\frac{{m}}{{s}}}}{{{\left({3.2}\times{10}^{{-{19}}}{C}\times{B}\right)}}}}\)

B = 2.02Tesla

b) Since current is measured in coulombs per second, the totalcharge moved per hour is the mass per hour over the mass per ion(which gives us the number of ions per hour), multiplied by thecharge per ion. This value is then divided by 3600seconds (perhour) to attain the current.

(massCollected / massPerIon) * CoulombPerIon / TimePeriod= Current

\(\displaystyle{\frac{{{\left\lbrace{.1}\times{10}^{{-{3}}}{k}{g}\right\rbrace}{\left\lbrace{3.92}\times{10}^{{-{25}}}{k}{g}\right\rbrace}}}{\times}}{\frac{{{\left\lbrace{3}\times{10}^{{-{19}}}{C}\right\rbrace}{\left\lbrace{3600}{s}\right\rbrace}}}{=}}{I}\)

I = .0227Amps

c) Energy dissipated in the cup is just a function of the kineticenergy dissipated for each ion times the number of ions collectedduring the period.

PSK\frac{1}{2}M(Velocity)^2 \times (massCollected / massPerIon)= EnergyDisspated \frac{1}{2} \times \frac{1}{2} \times (3.92\times 10^{-25}kg) \times (4.04\times 10^{5})^{2} \times \frac{.1\times 10^{-3}kg}{3.92\times 10^{-25}kg}) = EnergyZSK

Energy = \(\displaystyle{8.16}\times{10}^{{{9}}}{J}{o}\underline{{e}}{s}\)