What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56104V/m and a magnetic field of 4.6210-3T, with both fields normal to the beam and to each other, produces no deflection of the electrons?

Question

Mooney

Physics

12 January, 08:46

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56104V/m and a magnetic field of 4.6210-3T, with both fields normal to the beam and to each other, produces no deflection of the electrons?

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Answers (1)

Know the Answer?

Pierce Bailey · Answer 1

3.38 * 10^4 m/s.

Explanation:

Given:

E = 1.56*104V/m

B = 4.62*10-3T

F = q * V * B

E = F/q

E * q = q * V * B

V = E/B

= 1.56 * 10^4/4.62*10-3

= 3.38 * 10^4 m/s.

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56*104V/m and a magnetic field of 4.62*10-3T, with both fields normal to the beam and to each other, produces no deflection of the electrons?

What is the speed of a beam of electrons when the simultaneous influence of an electric field of 1.56104V/m and a magnetic field of 4.6210-3T, with both fields normal to the beam and to each other, produces no deflection of the electrons?