At some moment, an electron ( $q = - 1.6 \times 10 ^ { - 19 }$ C) moves with a velocity of 1 m/s orthogonal to a 1-T magnetic field. The force acting on the electron is
A) $0$ N.
B) $1.6 \times 10 ^ { - 19 }$ N.
C) $1.0$ N.
D) $1.0 \times 10 ^ { - 9 }$ N.

Question

Quizplus · Accepted Answer

The force acting on a charged particle moving in a magnetic field is given by $F = qvB\sin(\theta)$, where $q$ is the charge, $v$ is the velocity, $B$ is the magnetic field strength, and $\theta$ is the angle between the velocity and the magnetic field. Here, $\theta = 90^\circ$, so $\sin(\theta) = 1$. Substituting the given values: $F = (-1.6 \times 10^{-19} \, \text{C})(1 \, \text{m/s})(1 \, \text{T}) = -1.6 \times 10^{-19} \, \text{N}$. The magnitude of this force is $1.6 \times 10^{-19} \, \text{N}$, making option B correct.

At Some Moment, an Electron q=−1.6×10−19q = - 1.6 \times 10 ^ { - 19 }q=−1.6×10−19

At Some Moment, an Electron $q = - 1.6 \times 10 ^ { - 19 }$