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An electron travels at a speed of 16,748.76 m/s through a uniform magnetic field whose magnitude is 0.0177 T. What is the magnitude of the magnetic force on the electron if its velocity vector and the magnetic field vector are make an angle of 59.24° ?

Respuesta :

Answer:

Magnetic force, [tex]F=4.07\times 10^{-17}\ N[/tex]

Explanation:

It is given that,

Speed of electron, v = 16748.76 m/s

Magnetic field, B = 0.0177 T

Angle between velocity vector and the magnetic field vector are make an angle of 59.24°. Magnetic force is given by :

[tex]F=qvB\ sin\theta[/tex]

[tex]F=1.6\times 10^{-19}\times 16748.76\times 0.0177 \ sin(59.24)[/tex]

[tex]F=4.07\times 10^{-17}\ N[/tex]

So, the magnetic force on the electron is [tex]4.07\times 10^{-17}\ N[/tex]. Hence, this is the required solution.