60.12 m/s
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A particle of mass 8.25 kg is trapped in a magnetic field of strength 8.59T going at a speed of 60.12 m/s in a perfect circle of radius 5.80m as shown below. What is the charge on the particle? Ignore any gravitational effects.
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- An electron encounters an E and B fields.B field is given by B = 0.1 j T.Electron experiences a force F = (9.6 x 10^-14 i – 9.6 x 10^-14 k)Find the electric field encountered by the electron. The velocity of the electron isv = 5 x 10^6 i m/s.A proton is projected perpendicularly into a magnetic field that has a magnitude of 0.23 T. The field is then adjusted so that an electron will follow a circular path of the same radius when it is projected perpendicularly into the field with the same velocity that the proton had. What is the magnitude of the field used for the electron? Be=i. Find the direction of electric field, if any, at points P and Q due to the two charges as shown (two separate questions, spacings are uniform) -q Q I ii. Find the direction of magnetic force on electron at point R, and the direction of force on a proton at point S (two separate questions). Both the electron and proton has initial velocity upward (assume infinitely long current carrying wires). Avel S +q Р P vel e