A proton moving freely in a circular path perpendicularto a constant magnetic fi

Question

A proton moving freely in a circular path perpendicularto a constant magnetic field takes 1.90 µs to complete one revolution.Determine the magnitude of the magnetic field. I dont really know where to begin here. I pretty much have noclue what to do, i know the mass and charge on a proton, the anglewould be 90 degrees since its perpendicular. any suggestions on howto get started? A proton moving freely in a circular path perpendicularto a constant magnetic field takes 1.90 µs to complete one revolution.Determine the magnitude of the magnetic field. I dont really know where to begin here. I pretty much have noclue what to do, i know the mass and charge on a proton, the anglewould be 90 degrees since its perpendicular. any suggestions on howto get started?

Explanation / Answer

Magnetic field B = m / q                          = 2 m / q T where m = mass = 1.67 * 10 -27 kg            q= charge = 1.6 * 10 -19 C            T= 1.9 * 10 -6 s substitue values weget B = 3.5 * 10 -2T

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