The magnetic field over a certain range is given by B~ = Bx ˆı + By ˆ, where Bx = 2 T and By = 4 T. An electron moves into the field with a velocity ~v = vx ˆı+vy ˆ +vz ˆk, where vx = 2 m/s, vy = 6 m/s and vz = 8 m/s. The charge on the electron is - 1.602 * 10-19 C. What is the ˆı component of the force exerted on the electron by the magnetic field? Answer in units of N.

The force exerted in a magnetic field is given as

F = q (v * B)

Where

F is the force entered

q is the charge

v is the velocity

B is the magnetic field

Given that,

The magnetic field is

B = 2•i + 4•j. T

The velocity of the electron is

v = 2•i + 6•j + 8•k. m/s

Also, the charge of an electron is

q = - 1.602 * 10^-19 C.

Then note that,

V*B is the cross product of the speed and the magnetic field

Then,

F = q (V*B)

F = - 1.602 * 10^-19 (2•i + 4•j + 8•k * 2•i + 4•j)

Note

i*i=j*j*k*k=0

i*j=k. j*i=-k

j*k=i. k*j=-i

k*i=j. i*k=-j

F = - 1.602 * 10^-19[ (2•i + 4•j + 8•k) * (2•i + 4•j) ]

F = - 1.602 * 10^-19 [2*2• (i*i) + 2*4• (i*j) + 4*2• (j*i) + 4*4• (j*j) + 8*2• (k*i) + 8*4• (k*j) ]

F = - 1.602 * 10^-19[4•0 + 8•k + 8•-k + 16•0 + 16•j + 32•-i]

F = - 1.602 * 10^-19 (0 + 8•k - 8•k + 0 + 16•j - 32•i)

F = - 1.602 * 10^-19 (16•j - 32•i)

F = - 1.602 * 10^-19 * (-32•i + 16•j)

F = 5.126 * 10^-18 •i - 2.563 * 10^-18 •j

Then, the x component of the force is

Fx = 5.126 * 10^-18 N

Also, the y component of the force is

Fy = - 2.563 * 10^-18 N