Question

Two positive charges $q_1$ and $q_2$ are moving with velocities $v_1$ and $v_2$ when they are at points $A$ and $B$, respectively, as shown in figure. The magnetic force experienced by charge $q_1$ due to the other charge $q_2$ is :

• A. $\frac{{\mu }_0{q}_1{q}_2{v}_1{v}_2}{8\sqrt{2}\pi a^2}$
• B. $\frac{{\mu }_0{q}_1{q}_2{v}_1{v}_2}{4\sqrt{2}\pi a^2}$
• C. $\frac{{\mu }_0{q}_1{q}_2{v}_1{v}_2}{2\sqrt{2}\pi a^2}$
• D. $\frac{{\mu }_0{q}_1{q}_2{v}_1{v}_2}{\sqrt{2}\pi a^2}$

Answer 1

The correct option is A $\frac{{\mu }_0{q}_1{q}_2{v}_1{v}_2}{8\sqrt{2}\pi a^2}$

Magnetic field at $A$ due to $q_2$:
$\overrightarrow{B_2}=\frac{{\mu }_0}{4\pi }\frac{q_2{v}_{2}r\sin {45}^o}{r^3}\hat{k}$ where $r=\sqrt{2}a$ where $q_2{v}_2$ is the current due to motion of $q_2$
$\Rightarrow \overrightarrow{B_2}=\frac{{\mu }_0}{4\pi }\frac{q_2{v}_2}{2\sqrt{2}{a}^2}\hat{k}$
Force on $q_1:\overrightarrow{F}=q_1\overrightarrow{v_1}\times \overrightarrow{B_2}=\frac{{\mu }_0}{4\pi }\frac{{q_{1}q}_2{v}_1{v}_2}{2\sqrt{2}{a}^2}(-\hat{j})$