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

`a=2.6x10^(18) m/s^2`

Step-by-step 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`