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After a nuclear explosion an Uranium s2U235 nucleus is being released with a speed 1*10 ms As you can see from the symbol it has a mass 235 times the mass of a proton and positive charge 92 times in magnitude that of an electron. a) (10 points) Find the acceleration of the Uranium nucleus when its under the influence of Earths magnetic field which has a magnitude 10 and acting perpendicular to the b) (7 points) If this force acts for 15 seconds find the final velocity of the Uranium nucleus c) (5 points) If the final velocity of the nucleus and the acting magnetic force make the direction of the motion. after 15 s. Uranium nucleus to travel in a circular path find the radius of such a path d) (3 points) Now suppose the Uranium Nucleus is moving from left to right. And the magnetic field is pointed into the page as shown in the diagram. Use right hand rule to determine the direction of the magnetic force on the Uranium nucleus. Draw a rough sketch of the circular path traversed by the Uranium.
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a) The magnetic force acting on the Uranium nucleus is From the Newtons secondlaw From the above two equations dv M2 (10x10 T)(92x1.60x10 19 C)(10* m/s) (235x 1.67x10 kg) -27 = 3.75×106 m/s2 b) The final velocity of Uranium nucleus is y-y4at = (104 m/s)-(3.75x 106 m/s2) (15 s) = 5.63×10+7 m/s The radius of circular path is 235x1.67x10 k)(5.63x10/150*10 m 10x106 T 92x1.60x10-19 C d) From the right hand rule, the direction of magnetic force is upward

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