Answer to Question 2

An electron accelerated to 6.0 ´ 10^{6} m/s is deflected by a magnetic field of strength 0.82 T. It makes an angle of 76^{o} to the field.

(a) Calculate the force acting on the electron now.

(b) Calculate the forward velocity as it makes its helical path along the magnetic field.

(a) F = 0.82 T × 1.6 × 10^{-19} C × 6.0 × 10^{6} m s^{-1} × sin 76 = 7.64 × 10^{-13} N = 7.6 × 10^{-13} N (2 s.f.)

(b) v = 6.0 × 10^{6} m s^{-1} × cos 76 = 1.45 × 10^{6 }m s^{-1} = 1.5 × 10^{6} m s^{-1} (2 s.f.)