Question #158045

An electron in a television lamp moves in front of the tube at high speed

8. 10 ^ 6 m / s along the x-axis as shown. Magnetic fields create in

The tube is fitted with an angle of 60 ° to the x-axis

(in the xy plane) and

value 0. 025T.

a) Calculate the applied magnetic force and the acceleration of the electron.

b) Find rotation period, orbital radius and spiral step.


1
Expert's answer
2021-01-27T14:29:59-0500

a) F=qvBsinα=1.6101981060.025sin60°=2.771014 (T)F=|q|vB\sin\alpha=1.6\cdot10^{-19}\cdot8\cdot10^6\cdot0.025\cdot\sin60°=2.77\cdot10^{-14}\ (T)


a=F/m=2.771014/9.11031=31016 (m/s2)a=F/m=2.77\cdot10^{-14}/9.1\cdot10^{-31}=3\cdot10^{16}\ (m/s^2)



b) qvBsin60°=m(vsin60)2/rqvB\sin60°=m(v\cdot\sin60)^2/r and T=2πr/(vsin60°)T=2\pi r/(v\cdot \sin60°)\to T=2πmqB=T=\frac{2\pi m}{qB}=


=23.149.110311.610190.025=1.43109 (s)=\frac{2\cdot3.14\cdot 9.1\cdot10^{-31}}{1.6\cdot10^{-19}\cdot0.025}=1.43\cdot10^{-9}\ (s)



r=vmqBsin60°=81069.110311.610190.025sin60°=0.0016 (m)r=\frac{vm}{qB}\cdot\sin60°=\frac{8\cdot 10^6\cdot 9.1\cdot10^{-31}}{1.6\cdot10^{-19}\cdot0.025}\cdot\sin60°=0.0016\ (m)



h=vcos60°T=8106cos60°1.43109=0.00572 (m)h=v\cdot\cos60°\cdot T=8\cdot10^6\cdot\cos60°\cdot1.43\cdot10^{-9}=0.00572\ (m)











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