Question

An electron moves with speed 2*10^5ms along the positive x direction

Answer 1

Hey Dear,

# Complete question -

Q. An electron is moving with speed 2x10^5 m/s along the positive x-direction in the presence of magnetic induction B = i+4j-3k. Find magnitude of the force experienced by the electron.

◆ Answer -

|F| = 1.6×10^-13 N

◆ Explanation -

# Given -

q = 1.6×10^-19 C

v = 2×10^5 i m/s

B = i+4j-3k T

# Solution -

Magnitude of force experienced by electron is calculated by formula -

F = q.(v×B)

F = 1.6×10^-19 [(2×10^5 i)×(i+4j-3k)]

F = 1.6×10^-19 (6×10^5 j + 8×10^5 k)

F = 9.6×10^-14 j + 12.8×10^-14 k N

Magnitude of force is calculated by -

|F| = √[(9.6×10^-14)^2 + (12.8×10^-14)^2]

|F| = √(2.56×10^-26)

|F| = 1.6×10^-13 N

Therefore, electron will experience force of 1.6×10^-13 N.

Thanks dear...