Question

A proton with kinetic energy $1.3384 \times 10^{-14 \, \text{J}$ moving horizontally from north to south, enters a uniform magnetic field $B = 2.0 \, \text{mT}$ directed eastward. Calculate:}
\textbf{(a) the speed of the proton,

Hint:

For a charged particle in a magnetic field, use $F = qvB$ to calculate force, $a = F/m$ for acceleration, and $r = \frac{mv}{qB}$ for the radius of the path.

Answer 1

\normalfont a) The velocity \(v\) is given by the equation: \[ v = \sqrt{\frac{2 \times \text{K.E.}}{m}} \] Substituting the values: \[ v = 4 \times 10^6 \, \text{m/s} \] b) The acceleration \(a\) is given by the equation: \[ a = \frac{qvB}{m} \] Substituting the values: \[ a = 8 \times 10^{11} \, \text{m/s}^2 \] c) The radius \(r\) of the path is given by: \[ r = \frac{mv}{Bq} \] Substituting the values: \[ r = 20 \, \text{m} \]