Question

A constantly moving positively charged particle is spotted at point $(5\text{cm},0,0)$ at a given instant. At the same instant magnetic field produced by it at origin is detected to be $\overrightarrow{B}=B_0\hat{j}$; where $B_0$ is a positive constant. The possible instantaneous velocity of the charged particle is :

• A. $\overrightarrow{v}=5\hat{i}+4\hat{j}\text{m/s}$
• B. $\overrightarrow{v}=2\hat{i}-4\hat{k}\text{m/s}$
• C. $\overrightarrow{v}=-\hat{i}-3\hat{k}\text{m/s}$
• D. $\overrightarrow{v}=-2\hat{j}-2\hat{k}\text{m/s}$

Answer 1

Answer: (B), (C)

$\overrightarrow{v}=-\hat{i}-3\hat{k}\text{m/s}$


To produce magnetic field along $y$ direction, particle should move only in $x-z$ plane.

This , helps us exclude options (a) and (d).

From the data given in the question, instantaneous position of particle is $\overrightarrow{r}=5\hat{i}$

Let us consider $\overrightarrow{v}=x\hat{i}+z\hat{k}$

By applying Biot-Savert's law,

$\overrightarrow{B}=\frac{{\mu }_0}{4\pi r^2}(q\overrightarrow{v}\times \overrightarrow{r})$

$\Rightarrow B_0\hat{j}=\frac{{\mu }_0}{4\pi r^2}(q(x\hat{i}+z\hat{k})\times 5\hat{i})$

$\Rightarrow z=-\frac{4\pi r^2{B}_0}{5q{\mu }_0}$

$x$ can be either positive or negative, whereas $z$ can only be negative from the data obtained above.

Hence, options (b) and (c) are the correct answers.