Question

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)

Answer 1

Answer:

Explanation:

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