A proton with velocity ( 5x105 m/s) in +y direction, enters a region where B ( 0

Question

A proton with velocity ( 5x105 m/s) in +y direction, enters a region where B ( 0.1 T) is in +z direction. Calculate the radius of resulting circular path. Use [5].

What must have been the accelerating voltage, if above mentioned proton motion were a part of the mass-spectrometer (consisting parallel plate capacitor to accelerate charged ions to some velocity v, and then upon entering perpendicular B field the charges moved in circular trajectories. By adjusting the B field one could keep the radius constant, and monitor the current at fixed position, produced by charges entering the detector)? How strong a B is needed if the detected ion is charged iron (Fe+) instead? Use [6], and then use [7].

Explanation / Answer

Velocity v = 5 x 10 5 m / s Charge q = 1.6 x 10 -19 C Mass m = 1.67 x 10 -27 kg Magnetic field B = 0.1 T In magnetic field , Bvq = mv 2 / r                               Bq = mv / r From this radius r = mv / Bq Substitue values we get r = 0.05218 m (b). Potential difference V = [ ( 1/ 2) mv 2] / q       Since Vq = ( 1/ 2) mv 2                                         = 1304.68 volt

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