Answer:
19080667.0818 m/s
0.637294 m
[tex]2.1875\times 10^{15}[/tex]
Explanation:
m = Mass of deuterons = [tex]3.34\times 10^{-27}\ kg[/tex]
v = Velocity
K = Kinetic energy = 3.8 MeV
d = Diameter
B = Magnetic field = 1.25 T
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
t = Time = 1 s
i = Current = 350 μA
Kinetic energy is given by
[tex]K=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2K}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 3.8\times 10^6\times 1.6\times 10^{-19}}{3.34\times 10^{-27}}}\\\Rightarrow v=19080667.0818\ m/s[/tex]
The speed of the deuterons when they exit is 19080667.0818 m/s
In this system the centripetal and magnetic force will balance each other
[tex]\dfrac{mv^2}{r}=qvB\\\Rightarrow \dfrac{mv^2}{\dfrac{d}{2}}=qvB\\\Rightarrow d=\dfrac{2mv}{qB}\\\Rightarrow d=\dfrac{2\times 3.34\times 10^{-27}\times 19080667.0818}{1.6\times 10^{-19}\times 1.25}\\\Rightarrow d=0.637294\ m[/tex]
The diameter is 0.637294 m
Current is given by
[tex]i=\dfrac{nq}{t}\\\Rightarrow n=\dfrac{it}{q}\\\Rightarrow n=\dfrac{350\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\Rightarrow n=2.1875\times 10^{15}[/tex]
The number of deuterons is [tex]2.1875\times 10^{15}[/tex]
