Question

A particle of mass 6.5 times 10^-8 kg and charge +9.7 mu C is travelling due east. It enters perpendicularly a magnetic field whose magnitude is 4.4 T. After entering the field, the particle completes one-half of a circle and exits the travelling due west. How much time does the particle spend in the magnetic field? Number Units the tolerance is +/-5%

Answer 1

Mass of particle, $m = 6.5 \times 10^{-8} kg$

Charge of particle, $q =9.7 \times 10^{-6} C$

Existing Magnetic Field,

If the particle travels a semi-circular path, then distance travelled by it is

   $s= \pi r$

where radius of the path is given by

$r= \frac{mv}{qB}$

where v= speed of particle

Then time spent will be $t= \frac{s}{v}$

   $t= \frac{\pi r}{v}= \frac{\pi mv}{vqB}= \frac{\pi m}{qB}$

using given values in above,

   $t= \frac{3.14 \times 6.5 \times 10^{-8}}{9.7 \times 10^{-6}\times 4.4}= 4.78 \times 10^{-3}s$ (ANS)