Answer to Question #316601 in Electric Circuits for help

Question

A particle with mass 1.81 x 10-3 kg and a charge of 1.22 x 10-8 C has, at a given instant, a velocity 𝑣𝑣⃑ = (3.00 × 104 𝑚𝑚 𝑠𝑠 )𝚥𝚥̂. What are the magnitude and direction of the particle’s acceleration produced by a uniform magnetic field 𝐵𝐵�⃑ = (1.63 𝑇𝑇)𝚤𝚤̂+ (0.980 𝑇𝑇)𝚥𝚥̂̂?

Answer 1

The magnetic field force

$$\vec F=q\vec v\times \vec B$$

$$\vec F=1.22*10^{-8}*(3.00\times 10^4\hat j)\times (1.63\hat i+0.980\hat j)\\ =-5.97*10^{-4}\hat k$$

The acceleration

$$\vec a=\frac{\vec F}{m}=\frac{-5.97*10^{-4}\hat k}{1.81*10^{-3}}=-0.33\hat k\:\rm m/s^2$$