Charged Particle Moving in a Magnetic Field

Hello. Can you please check out the attachment?

Solving (a) is easy. The correct answer to (b) is "the period for the return trip is unchanged" (that is, T = 130 ns).

Okay, fine: the attachment shows why T only depends on m, q, and B.

How on Earth, though, do you reconcile this with the inverse relationship between velocity and period, given the initial and return arc radius values are the same???

In other words, if KE is doubled, the speed increases, and (again for a constant radius) the period must DECREASE!

What am I missing? Thank you!

Attached Images
File Type: jpg Capture.JPG (84.2 KB)

http://ift.tt/SGKwkp

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