When an electron enters a region of uniform magnetic field (B=0.19T), with its
velocity (V=5.5X10^6m/s) perpendicular to the field, it experiences a magnetic
force which acts as centripetal force to cause circular motion of the particle. find
the radius of the track?

When an electron enters a region of uniform magnetic field B019T with its velocity V55X106ms perpendicular to the field it experiences a magnetic force which ac class=

Respuesta :

The radius of the track in the circular motion is 0.164 mm.

Magnetic force of the of charge

The magnetic force of the charge is calculated as follows;

F = qvB

Centripetal force of the charge

The centripetal force experienced by the charge is calculated as follows;

[tex]F = \frac{mv^2}{r}[/tex]

The magnetic force is equal to the centripetal force

[tex]qvB = \frac{mv^2}{r} \\\\qB = \frac{mv}{r} \\\\r = \frac{mv}{qB} \\\\[/tex]

The radius of the track

The radius of the track is calculated as follows;

[tex]r = \frac{9.1 \times 10^{-31} \times 5.5 \times 10^6}{1.6 \times 10^{-19} \times 0.19} \\\\r = 1.64 \times 10^{-4} \ m\\\\r = 0.164 \times 10^{-3} \ m\\\\r = 0.164 \ mm[/tex]

Thus, the radius of the track in the circular motion is 0.164 mm.

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