An electron moves within a uniform magnetic field. its instantaneous velocity is v = vy j + vz k, where vy =5m/sandvz =3m/s. thefieldisgivenbyb=bxi+byjwherebx =2tandby =4t. thechargeonthe electron is −1.6 × 10−19 c. what force f is exerted on the electron by the magnetic field?

F = 3.2 x 10⁻¹⁹ x (6i - 3j + 5k) N

Explanation:

Magnetic fields exert forces on moving charge. The force F exerted by a  magnetic field B on a charge q moving with velocity v in an electric field is called the magnetic Lorentz force. It is given by

                                            F = qv × B  

The magnitude of the Lorentz force F is F = qvB sinθ, where θ is the smallest angle between the directions of the vectors v and magnetic field B. When  v and B are perpendicular to each other, sinθ = 1 then F has its maximum possible magnitude F = qv x B

Given:

Instantaneous velocity v = vyj + vzk   but  vy =5m/s and vz =3m/s.

                                           v = 5j + 3k   m/s

Field B = Bxj + Byj  but  Bx =2T and By =4T

                                            B = 2i + 4j  T

From magnetic Lorentz force F = q v x B

                              F = −1.6 × 10⁻¹⁹ x [ (5j + 3k) X (2i + 4j) ]  

solving vector multiplication of V X B

                          =

                          V X B =

                           F = −1.6 × 10⁻¹⁹ x (-12i + 6j -10k)

Expanding the bracket and simplifying to the lowest term;

                           F = 3.2 x 10⁻¹⁹ x (6i - 3j + 5k) N

0.19 m/s and 0.45 m/s