Question

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A uniform magnetic field of magnitude 3.00 LT is applied in the negative x direction. An electron is shot at the origin with an initial velocity of v (1.50 i+ 10.0 k) x 10° m/s and moves in a helix, as shown. Determine the electron's acceleration in the x direction. а. the radius of the helical path. b. the time is takes the electron to complete one revolution. c. В

Answer 1

​​​​​​By balancing centripetal force and magnetic force we may draw important results.

to 9 r TTUU EB eriven magnetic field = 3.00 MT = 3X66T velocity of o = (1.50 +1000 ) xos misec @ The acceleration of charge particle is only due to electric field. There is no effect of magnetic field on its relocity. The magnetic field only give a direction of charged partide here circular motion So electron's acceleration in x-direction is zero. acceleratino 0 Heka v = le.o Rхь mla el is arcadas m-ho equating magnetic force = centrifugal force Bqv = mi to r= mus B Bre k= mass of elechon x velocity along 3 direction magnetic field x charge on electron = 9.181031X 106+6 =1.895 x 10° . =1.9m radiis, . Time = axr = 21 x mu Vk & T= eam - 8X314 x 9.1X1631 Bq 3 X106X16X1=19 = 11.9x10sec 1. Hence time taken to complete me revolution is 11.9 used - 3x156x 1.681019 VK C . . .