Question

A crossed-field velocity selector has a magnetic field of magnitude 0.03 T. The mass of the electron is 9.10939 × 10−31 kg. What electric field strength is required if 6.5 keV electrons are to pass through undeflected?

Answer 1

Answer:

E = 4.78*10^6 N/C

Explanation:

In this case you have to take into account that both magnetic force and electric force must be equal

$F_E=F_B\\\\qE=qvB$

$E=vB$

However, you have to know first what is the speed of the electron by using the expression for the kinetic energy (1keV=1000*1.6*10^{-19}J):

$E_k=\frac{1}{2}m_ev^2\\\\v=\sqrt{\frac{2E_k}{m_e}}=\sqrt{\frac{2(6500(1.6*10^{-19}J))}{9.1*10^{-31}kg}}=4.78*10^{7}\frac{m}{s}$

by replacing in the expression for the forces you have

$E=(4.78*10^{7}\frac{m}{s})(0.03T)=1.43*10^{6}N/C$

HOPE THIS HELPS!!