A proton traveling at 319 degree with respect to the directi

A proton traveling at 31.9 degree with respect to the direction of a magnetic field of strength 2.78 mT experiences a magnetic force of 5.13 times 10^-17 N. Calculate (a) the proton\'s speed and (b) its kinetic energy in electron-volts. Number Units Number Units

Solution

here ,

theta = 31.9 degree

magnetic field , B = 2.78 mT

force , F = 5.13 *10^-17 N

a) let the proton\'s speed is v

F = q *V * B * sin(theta)

5.13 *10^-17 = 0.00278 * sin(31.9 degree) * 1.602 *10^-19 * v

v = 2.18 *10^5 m/s

the speed of the proton is 2.18 *10^5 m/s

b)

Now , kinetic energy = 0.5 * m * v^2/charge

kinetic energy = 0.5 * 1.67 *10^-27 * (2.18 *10^5)^2/(1.602 *10^-19)

kinetic energy = 247.7 eV = 0.2477 keV