Q2 [5 marks]. Consider a proton moving across an equator of the Earth towards 35° NE-direction, at a speed of 2500 km/s. If magnetic field at a crossing point is towards 90° direction relative to equator, and has a magnitude of about 0.5 micro-Tesla, how much force would be exerted on a proton?
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![Q2 [5 marks]. Consider a proton moving
across an equator of the Earth towards
35° NE-direction, at a speed of 2500
km/s. If magnetic field at a crossing point
is towards 90° direction relative to
equator, and has a magnitude of about
0.5 micro-Tesla, how much force would
be exerted on a proton?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff6b66fd0-f30c-421b-800d-5229914c807e%2Ff5cb08c9-4583-4437-b18f-f5132953da5e%2F0fisvt3i_processed.jpeg&w=3840&q=75)
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- 8. In an experiment with cosmic rays, a vertical beam of particles that have charge of magnitude 3e and mass 12 times the proton mass enters a uniform horizontal magnetic field of 0.250 T and is bent in a semicircle of diameter 95.0 cm. Find the my speed of the particles and the sign of their charge. (Hint: R= Magnitude of charge of electron e = 1.6021765651352 * 10" C, Mass of proton m₂ = 1.6726217771742 * 10 kg) (10') 98² 95.0 cmAn electron moves at a speed of 810 m/s perpendicular to the direction of a uniform magnetic field of 0.4 T. What is the radius of the electron's circular orbit in units of nanometers (1 m = 10° nm)? (in units of nanometers)help