AL electromagnetism question

2011-10-28 2:53 am

(When a proton of negligible speed is injected into the centre of the device, it is accelerated by the p.d. towards one dee and describes a semi - circle inside that dee due to the magnetic field present. The proton is then accelerated across the gap through a p.d. of V and describes another semi - circle of greater radius in the other dee since its speed has increased. This process is repeated and the path of proton is shown in the figure.)

http://i213.photobucket.com/albums/cc101/Glamorous_Premonition/phy.jpg

Question: It is known that the strength of the magnetic field , B , is 1.5 T and the alternating p.d. , V , is 10 k V.(Given : electronic charge = 1.60 x 10-19 C ; mass of proton = 1.66 x 10-27 kg )(i) Find the gain in kinetic energy , in eV , of the proton in each complete revolution

the answer : 2 2(e x 10 kV)= 20keV

I don't understand why 2 is needed and the direction of B-field.
Please provide with detail explanation. Thank you very much!

回答 (1)

天同

2011-10-28 3:35 am

✔ 最佳答案

The device is a cyclotron.

In a complete revolution, the proton passes through the open gap TWO times. Each time of passing, the proton acquire energy of eV ( V = 10 kV), hence in two passings, the proton acquires energy of 2eV

The magnetic force acting on the proton provides the necessary centripetal force for the proton to revolve round a circle. Given the direction of proton as shown on the figure, it is not difficult, by using Fleming's Left Hand Rule, to find that the magnetic field lines are pointing into the paper.

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