Consider the mass spectrometer shown schematically below: The electric field bet

Question

Consider the mass spectrometer shown schematically below: The electric field between the plates of the velocity selector is 2500 v/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.0350 T, Calculate the radious of the path of a singly ion having a mass = 2.18 times 10^-26 kg. what is the required radius of a cyclofrom designed for accelarate protons to energies of 34.0 MeV using a magnetic field of 5.20 T? A proton moves in a direction perpendicular to a uniform magnetic field B rightarrow at 1.00 times 10^7 m/s and experiences an accelaration of 2.00 times 10^13 m/s^2 in the +x direction. when its velocity is in the +z direction. Determine the magnitude and direction of the field.

Explanation / Answer

To do this, we use the following two expressions:
K = 1/2 mv2 ---> solving for v:
v = (2K/m)1/2 (1)
K = 34 MeV * 1.602x10-13 J/MeV = 5.447x10-12 J

And this expression:
F = qvB ---> F = ma
ma = qvB   ---> a = v2/r
mv2/r = qvB
r = mv/qB   (2)

Replacing 1 in 2:
r = (2Km)1/2 / qB

Now replacing the values we have:
r = (2 * 5.447x10-12 * 1.67x10-27)1/2 / 1.602x10-19 * 5.2
r = 0.1619 m or 16.19 cm

Hope this helps

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