A small particle accelerator for accelerating 3He+ ions is shown in the figure.

Question

A small particle accelerator for accelerating 3He+ ions is shown in the figure. The 3He+ ions exit the ion source with a kinetic energy of 5 keV. Regions 1 and 2 contain magnetic fields directed into the page, and region 3 contains an electric field directed from left to right. The 3He+ ion beam exits the accelerator from a hole on the right that is 7 cm below the ion source, as shown in the figure. If B1 = IT and region 3 is 50 cm long with E = 68 kV/m, what value should B2 have to cause the ions to move straight through the exit hole after being accelerated twice in region 3? What minimum width X should region 1 have? What is the velocity of the ions when they leave the accelerator?

Explanation / Answer

Mass of Helium-3 atom = 3*1.66 * 10^-27 = 4.98 * 10^-27 Kg

Initially ,

energy = 5 KeV

so, velocity = sqrt ( (2 * 5000 * 1.6 * 10^-19 )/ 4.98 * 10^-27 )

= 566721.566 m/sec

for first semi-circle

mv^2 / r = B * q * v

r1 = m * v / (B * q ) = 1.764 cm

After passing through region 3 , the ion will accelerate ,

energy of ion after passing through region 3 = 5 KeV + 34 KeV = 39 KeV

so , velocity =sqrt ( (2 * 39000 * 1.6 * 10^-19 )/ 4.98 * 10^-27 ) = 1583042.67 m/sec

for second second semi-circle ,

r2 = m * v / (B2 * q ) = 4.75 B2 cm

Now , the ion wont accelerate , so energy = 39 KeV

so , velocity =1583042.67 m/sec

for third second semi-circle ,

r3 = m * v / (B1 * q ) = 4.75 cm

so , velocity =sqrt ( (2 * 73000 * 1.6 * 10^-19 )/ 4.98 * 10^-27 ) = 2165816.948 m/sec

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