An electron traveling toward the north with speed 4.0 * 105 m/s enters a region where the Earth's magnetic field has the magnitude 5.0 * 10-5 T and is directed downward at 45° below horizontal. What is the magnitude of the force that the Earth's magnetic field exerts on the electron? (e = 1.60 * 10-19 C)

Speed of electron

v = 4 * 10^5 •j m/s

Magnetic field

B = 5 * 10^-5 T at angle of 45° to horizontal

Charge of electron

q = 1.6 * 10^-19C

Magnitude of force F?

The Force exerted in an electric field is given as

F = q (v*B)

Now, x component of the magnetic field

Bx = BCos45 = 5*10^-5 Cos45

Bx = 3.54 * 10^-5 •i T

Also, y component

By = BSin45 = 5 * 10^-5Sin45

By = 3.54 * 10^-5 •j T

B = 3.54 * 10^-5 •i + 3.54 * 10^-5 •j T

Now, F = q (v*B)

Note that,

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

i*j = k, j*k = i, k*i = j

j*i = - k, k*j = - i and i*k = - j

Therefore

F = q (v*B)

F = 1.6*10^-19 (4*10^5•j * (3.54 * 10^-5 •i + 3.54 * 10^-5 •j T))

F = 1.6*10^-19 (4*10^5 * 3.54 * 10^-5 (j*i) + 4*10^5 * 3.54 * 10^-5 (j*j))

F = 1.6*10^-19 (14.14 (-k) + 0)

F = - 2.26 * 10^-18 •k N

It is in the negative direction of z axis

The magnitude of the force the field experience is 2.26 * 10^-18 N