Question

A mass spectrograph is operated with deuterons, which have a charge of +e and a mass of 3.34 x 10-27 kg. Deuterons emerge from the source, which is grounded with negligible velocity. The velocity of the deuterons as they pass through the accelerator grid is 8.0 x 105 m/s. A uniform magnetic field of magnitude B = 0.20 T, directed out of the plane, is present at the right of the grid. In the figure, the deuterons are in circular orbit in the magnetic field. The radius of the orbit and the initial sense of deflection are closest to:

Answer 1

0.0835 m

Step-by-step explanation:

The magnetic force exerted on the deuteron is equal to the centripetal force that keeps it in circular motion:

`qvB = m(v^2)/(r)`

where

`q=+e = 1.6\cdot 10^(-19) C` is the charge

`v=8.0\cdot 10^5 m/s` is the speed of the deuteron

`B=0.20 T` is the magnetic field strength

`m=3.34\cdot 10^(-27) kg` is the mass of the deuteron

r is the radius of the orbit

re-arranging the equation, we find:

`r=(mv)/(qB)=((3.34\cdot 10^(-27)kg)(8.0\cdot 10^5 m/s))/((1.6\cdot 10^(-19) C)(0.20 T))=0.0835 m`