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A cosmic ray proton moving toward the earth at 5.03 x 10^7 m/s experiences a magnetic force of 1.68 x 10^16 N. What is the strength of the magnetic field if there is a 42-degree angle between it and the protons velocity?
A 2.8 C charge with a velocity v = 2i +0j + 3k m/s enters a region where there is a uniform magnetic field of B = 2i + 3j + 1k; compute the vector force on the charge.
A DC power line for a light-rail system carries 900 A at an angle of 29.5 degrees to the earth's 4.96 x 10^5 T field. What is the force on a 100m section of this line?
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A proton moves at 7.51 x 10^7 m/s perpendicular to a magnetic field. The field causes the proton to travel in a circular path of a radius of 0.796 m. What is the field strength?
A.) At what speed will a proton move in a circular path of 4.25 m where the field strength is 0.96 x 10^5 T?
B.) What would the radius of the path be if the proton had a speed of 7.4 x 10^6 m/s?
When the electric field is turned off, the charged particle travels in a circular path of radius 10mm, as it travels through the magnetic field (still at B=0.6T). What is the mass-charge ratio of the particle?
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