Question

A negatively charged particle (charge = -1e) has a velocity of 6.0 × 106 m/s in the positive x direction at a point where the magnetic field has the components Bx = 3.0 T, By = 1.5 T, and Bz = 2.0 T. What is the magnitude of the acceleration of the particle?

Answer 1

Answer:

$a=2.6x10^{18} m/s^2$

Explanation:

$v=6.0x10^6 m/s$, $q=1.6x10^{-19}C$, $q=9.11x10^{-31}kg$

$\beta=\beta_x=3.0T$, $\beta_y=1.5 T$, $\beta_y=2.0 T$

$F_b=q*\left[\begin{array}{ccc}i&j&k\\6x10^6&0&0\\3&1.5&2\end{array}\right]$

$F_b=-1.6x10^{-19}C*[-12x10^6 j + 9x10^6 k]$

$F_b=m*a$

$19.2x10^{-13}j-14.4x10^{-13}k=m*a$

$a=\frac{19.2x10^{-13}}{9.11x10{-31}}-\frac{14.4x10^{-13}}{9.11x10{-31}}$

$a=2.107 x10^{18}j-1.58x10^{18}k$

$a=\sqrt{(2.107x10^{18})^2-(1.57x10^{18})^2}$

$a=2.6x10^{18} m/s^2$