Question

Question 1 5 pts A charged particle, q - 4.00 pc, enters a uniform perpendicular magnetic field of strength 150.00 mt with an instantaneous speed of 5.00 x 10°/s and begins to circulate. Due to energy losses because of synchrotron radiation the radius of curvature begins to decrease, so the particle spirals inward. If the radius of curvature decreases at the rate of 0.05 mm/sec. determine the rate of energy loss in Joules/sec. HTML Editor BI U ANI EEx , E - SC 2 N G 12p . Paragraph -

Answer 1

solution: Given charged particle . Given charged particle , q = 4.00 UC - q = 4x10°C enters into magnetic field strength = 150.comT B= 150x1037 with instantaneous speed, V = 5 x 104 m/sec. Rate of decrease of Radius of curvature, 0.05 mm/sec. dx = 0.05x10°misec. situation is like , Now applying Law of centripetal force F = mv² - and in a uniform magnetic field we know, F = QUB - Hence, from 0 t @ que - mus Rate of change of Equation ☺ d = m. dve dt qB at 1: m, 9,8 = constant Vrthanges with change in radius) no

Now the kinetic Energy can be written as, E = KE = tmuz Now Rate of Energy loss is, de more Put the values of mdu = 98.dr from @ dt de Hence, Rate of Energy loss, Male = que de Now substituting all the given values, dE dt = 4.00X156 x 5.00X109 x 150.00X1630.05X173 dE - 150X106 Joule sec