Hi! I saw the answer of problem 27 in the chapter 19. Theanswer is said that mass of the particle is 2.0x10kg. But thequestion is asking about magnetic field. the question isbelow, "A proton moving freely in a circular path perpendicular to aconstant magnetic field taks 1.00s to complete onerevolution. Determin the magnitude of the magnetic field." I just want to make sure this is the right anawer for thequestion. Tamiko Hi! I saw the answer of problem 27 in the chapter 19. Theanswer is said that mass of the particle is 2.0x10kg. But thequestion is asking about magnetic field. the question isbelow, "A proton moving freely in a circular path perpendicular to aconstant magnetic field taks 1.00s to complete onerevolution. Determin the magnitude of the magnetic field." I just want to make sure this is the right anawer for thequestion. Tamiko
Explanation / Answer
time taken to complete one revolution T = 1 s = 1* 10 ^-6 s i.e., time period T = 10 ^ -6 s angular frequency w = 2 / T = 6.28 * 10^ 6 rad / s In magnetic field Bvq = mr w ^ 2 from this magnetic field B = mrw^ 2 / v q = mrw ^ 2 / rw q = mw / q = [ (1.67 * 10 ^ -27 kg ) ( 6.28 * 10 ^ 6 rad / s ) ] / [ 1.6 * 10^ -19 C ] = 6.55 * 10 ^ -2 T