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An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic field, oriented perpendicular to the electrons velocity. what is the magnitude of the magnetic force on the electron?

Respuesta :

Given :

Potential difference, V = 300 V.

Magnetic Field, B = 0.001 T.

To Find :

The magnitude of the magnetic force on the electron.

Solution :

We know, for perpendicular orientation, force is given by :

[tex]F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N[/tex]

Hence, this is the required solution.

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