Question

A particle of mass 6.3 x 10-8kg and charge 7.1 muC is traveling due east. It enters perpendicularly a magnetic field whose magnitude is 1.7 T. After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend traveling in the magnetic field

Answer 1

Answer:

t=0.016s

Explanation:

we can use

$\frac{v}{r}=\omega=\frac{qB}{m}=\frac{7.1*10^{-6}C*1.7T}{6.3*10^{-8}kg}=191.58s^{-1}$

the time that the particle is in the magnetic field is one half oa period. Hence

$t=\frac{T}{2}=\frac{\pi}{\omega}=0.016s$

I hope this is useful for you

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