(a) 0.579 T
When the ions are accelerated through the potential difference, the kinetic energy gained is equal to the change in electric potential energy:
[tex]\frac{1}{2}mv^2 = q\Delta V[/tex]
where m is the mass of the ions, v their final speed, q their charge, and [tex]\Delta V[/tex] the potential difference.
We can re-write the equation as
[tex]v=\sqrt{\frac{2q\Delta V}{m}}[/tex] (1)
When the ions enter the region with magnetic field, the magnetic force acts as centripetal force. So we can write
[tex]qvB = m\frac{v^2}{r}[/tex] (2)
where B is the magnitude of the magnetic field and r is the radius of the path. Combining (1) with (2), we find an expression for the magnetic field:
[tex]B=\frac{1}{r}\sqrt{\frac{2m\Delta V}{q}}[/tex]
And using:
m = 3.92 × 10-25 kg
q = 3.20 × 10-19 C
[tex]\Delta V=112 kV = 1.12\cdot 10^5 V[/tex]
r = 0.904 m
We find:
[tex]B=\frac{1}{0.904}\sqrt{\frac{2(3.92\cdot 10^{-25})(1.12\cdot 10^5)}{(3.20\cdot 10^{-19})}}=0.579 T[/tex]
(b) [tex]3.19\cdot 10^{-4}A[/tex]
The total mass separated per hour is
[tex]M = 1.41 mg = 1.41\cdot 10^{-6} kg[/tex]
Which corresponds to a number of ions of
[tex]n=\frac{M}{m}=\frac{1.41\cdot 10^{-6} kg}{3.92\cdot 10^{-25} kg}=3.60\cdot 10^{18}[/tex]
So the total charge of these ions is
[tex]Q=Nq = (3.60\cdot 10^{18})(3.20\cdot 10^{-19}=1.15 C[/tex]
This is the total charge passing through the slit in one hour, which corresponds to a time of
t = 1 h = 3600 s
Therefore, the current is
[tex]I=\frac{Q}{t}=\frac{1.15 C}{3600 s}=3.19\cdot 10^{-4}A[/tex]
(c) [tex]1.03\cdot 10^5 J[/tex]
First of all, we need to calculate the number of ions collected in the cup in 0.796 h.
We already know that the number of ions collected in 1 hour is (part b)
[tex]3.60\cdot 10^{18}[/tex], therefore
[tex]N' = (3.60\cdot 10^{18}) \cdot \frac{0.796 h}{1h}=2.87\cdot 10^{18}[/tex]
The kinetic energy carried by each ion is (part a)
[tex]E_k = q\Delta V=(3.20\cdot 10^{-19}C)(1.12\cdot 10^5 V)=3.58\cdot 10^{-14} J[/tex]
Therefore, the thermal energy produced in the cup in 0.796 h will be equal to the energy carried by a single ion times the number of ions:
[tex]E=N' E_k = (2.87\cdot 10^{18})(3.58\cdot 10^{-14} J)=1.03\cdot 10^5 J[/tex]
