Hello, Can anybody pleae help me out with this problem?I can\'t come up with the

Question

Hello,    Can anybody pleae help me out with this problem?I can't come up with the write answer and keep getting it markedwrong. Thanks. :) A proton moving freely in a circular pathperpendicular to a constant magnetic field takes 0.60 µs to complete one revolution. Determinethe magnitude of the magnetic field. Hello,    Can anybody pleae help me out with this problem?I can't come up with the write answer and keep getting it markedwrong. Thanks. :) A proton moving freely in a circular pathperpendicular to a constant magnetic field takes 0.60 µs to complete one revolution. Determinethe magnitude of the magnetic field.

Explanation / Answer

In magnetic field , force due to magnetic field = centripetalforce                                                         B v q = m r 2                                                      B r q = m r 2      Since   v = r from this angular frequency = B q / m from this magnetic field B = m / q where = 2 / T            T= period = 0.6 s              = 0.6 * 10 ^ -6 s            m= mass of proton = 1.67 * 10 ^ -27 kg            q = charge = 1.6 * 10 ^ -19 C substitue values we get answer                                                      B r q = m r 2      Since   v = r from this angular frequency = B q / m from this magnetic field B = m / q where = 2 / T            T= period = 0.6 s              = 0.6 * 10 ^ -6 s            m= mass of proton = 1.67 * 10 ^ -27 kg            q = charge = 1.6 * 10 ^ -19 C substitue values we get answer

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