JEE Main 2025 — 23 January, Evening Shift — Question 56

$B=\frac{\overline{\mu_{0} I}}{2 \pi r}$

$\Rightarrow\left[\mu_{0}\right]=\left[\frac{\mathrm{B} \times \mathrm{r}}{\mathrm{I}}\right]=\left[\frac{\mathrm{MT}^{-2} \mathrm{~A}^{-1} \times \mathrm{L}}{\mathrm{A}}\right]=\left[\mathrm{MLT}^{-2} \mathrm{~A}^{-2}\right]$

magnetic field

$\mathrm{F}=\mathrm{qvB}$

$\text{B}=\left[ \frac{\text{ML}{{\text{T}}^{-2}}}{\text{A L}}\right]=\left[ \text{M}{{\mathrm{T}}^{-2}}\text{ }{{\mathrm{A}}^{-1}} \right]$

$\text{M} = \text{I .A}$ $=[ \text{A} ]. [ \text{{L}}^{2}]=[ \text{{L}}^{2} . \text{A}]$

$=\text{c}\theta \Rightarrow \text{c}=\left[ \frac{\tau }{\theta } \right]=\left[ \mathrm{M}{{\mathrm{L}}^{2}}\mathrm{ }{{\mathrm{T}}^{-2}} \right]$