Question

A charged particle with velocity $\overrightarrow{v}=x\hat{i}+y\hat{j}$ moves in a magnetic field $\overrightarrow{B}=y\hat{i}+x\hat{j}$. The force acting on the particle has magnitude $F$. Which one of the following statements is/are correct?

• A. No force will act on charged particle if $x=y$
• B. If $x>y,F\propto (x^2-y^2)$
• C. If $x>y$, the force will act along z-axis
• D. If $y>x$, the force will act along y-axis

Answer 1

Answer: (A), (B), (C)

No force will act on charged particle if $x=y$
If $x>y,F\propto (x^2-y^2)$
If $x>y$, the force will act along z-axis

$\overrightarrow{v}=x\hat{i}+y\hat{j}$

$\overrightarrow{B}=y\hat{i}+x\hat{j}$

$\overrightarrow{f}=q(\overrightarrow{v}\times \overrightarrow{B})=q\left(\right(x\hat{i}+y\hat{j})\times (y\hat{i}+x\hat{j}\left)\right)=q(x^2-y^2)\hat{k}$

If $x=y$, $\overrightarrow{f}=q(x^2-y^2)=q(0-0)=0$

If $x>y$, $\overrightarrow{f}=q(x^2-y^2)\hat{k}>0$

If $y>x$, $\overrightarrow{f}=q(x^2-y^2)\hat{k}<0$ and it points in the $-vez-$ axis.