An electron enters a region of magnetic field of magnitude0.016 T, traveling per

Question

An electron enters a region of magnetic field of magnitude0.016 T, traveling perpendicular tothe linear boundary of the region. The direction of the field isperpendicular to the velocity of the electron. (a) Determine the time it takes for theelectron to leave the "field-filled" region, noting that its pathis a semicircle.
1 ns
(b) Find the kinetic energy of the electron if the radius of itssemicircular path is 2.00 cm.
2 keV (a) Determine the time it takes for theelectron to leave the "field-filled" region, noting that its pathis a semicircle.
1 ns
(b) Find the kinetic energy of the electron if the radius of itssemicircular path is 2.00 cm.
2 keV

Explanation / Answer

Hi, We have the radius of the circular path of charged particle in amagnetic field is r = mv/qB Then time taken by the charged particle to come out of the field ist = r/v                                                                                                     =m/qB                                                                                               =[()(9.11*10-31kg)]/[(1.6*10-19C)(0.016T)]                                                                                                =1118*10-12s                                                                                                =1.12ns kinetic energy K = (1/2)mv2                          = (1/2)m(rqB/m)2                                   = (rqB)2/2m                          =[(2*10-2m)(1.6*10-19C)(0.016T)]2/[(2)(9.11*10-31kg)]                          =1.439*10-15J                          =(1.439*10-15J)/(1.6*10-16J/keV)                          = 9 keV Hope thios helps you

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