Question

A charged particle of charge $2\text{mC}$ has a velocity equal to $2\text{m/s}$ along the positive $x-$axis. If in the region, there exists a uniform magnetic field of $4\text{T}$, directed along the negative $x-$axis. The magnitude of magnetic force on the charged particle is equal to

​​​​​​​$[$0.77 Mark$]$

• A. $2\text{N}$
• B. $4\text{N}$
• C. $6\text{N}$
• D. Zero

Answer 1

The correct option is D Zero

Force experienced by a charged particle moving in a magnetic field is given by,

$\left|\begin{array}{c}\overrightarrow{F}\end{array}\right|=q\left|\begin{array}{c}\overrightarrow{v}\times \overrightarrow{B}\end{array}\right|=qvB\sin \theta$

Where, $\theta$ is the angle between $\overrightarrow{v}$ and $\overrightarrow{B}$.

Here, $\overrightarrow{v}$ is along positive $x-$ axis and $\overrightarrow{B}$ is along negative $x-$axis.

So, $\theta ={180}^{\circ }\Rightarrow F=0$