Physics, asked by pruthvi4554, 10 months ago

An electron experiences a magnetic force of magnitude 4.60 xx 10^-15 N, when moving at an angle of 60^@ with respect to a magnetic field of magnitude 3.50 xx 10^-3 T. Find the speed of the electron.

Answers

Answered by minku8906
0

The speed of the electron is

Explanation:

Given :

Force F = 4.60 \times 10^{-15} N

Magnetic field B = 3.50 \times 10^{-3} T

According to force experience in a magnetic field,

   F = qvB \sin \theta

Where \theta = angle between velocity and magnetic field.

In our case angle between velocity and magnetic field is 60°

So speed of electron is,

    v = \frac{F}{qB \sin 60}                                       ( q = 1.6 \times 10^{-19} )

    v = \frac{4.6 \times 10^{-15} }{1.6 \times 10^{-19 \times 3.5 \times 10^{-3}  \times 0.866} }

    v = 9.48 \times 10^{6} \frac{m}{s}

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