Question

An electron moves with speed 2×10⁵ m/s along the positive x-direction in the presence of a magnetic induction B=^i+4^j−3^k (in Tesla.) The magnitude of the force experienced by the electron in Newton's is (charge on the electron = 1.6×10⁻¹⁹ C)

A. 1.18×10⁻¹³ N
B. 1.28×10⁻¹³ N
C. 1.6×10⁻¹³ N
D. 1.72×10⁻¹³ N

Answer 1

the magnitude of the force is 3.2 × 10^-14 N. The force experienced by the electron can be calculated using the equation F = qvBsinθ. In this case, the angle between the velocity and the magnetic field is 90 degrees or pi/2 radians.

Step-by-step explanation:

The force experienced by the electron can be calculated using the equation F = qvBsinθ, where F is the force, q is the charge of the electron, v is the velocity of the electron, B is the magnetic induction, and θ is the angle between the velocity and the magnetic field.

Since the electron is moving along the positive x-direction, the velocity vector is parallel to the x-axis.

This means that the angle between the velocity and the magnetic field, θ, is 90 degrees or pi/2 radians.

Plugging in the values given in the question, we have

F = (1.6×10^-19 C)(2×10^5 m/s)(1 T)sin(pi/2)

= (1.6×10^-19 C)(2×10^5 m/s)(1 T)(1)

= 3.2 × 10^-14 N.

Therefore, the magnitude of the force experienced by the electron is 3.2 × 10^-14 N.