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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?

Respuesta :

Answer:

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.

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