Question

The ratio of dimensions electric field and magnetic field is,

• A. $M{T}^1{L}^{-1}$
• B. $M{T}^{-1}{L}^{-2}$
• C. $M{T}^2{L}^{-1}$
• D. $L{T}^{-1}$

Answer 1

Answer: (D)

$L{T}^{-1}$

Electric field $=$ force/charge or $E=F/q$

So, $\left[E\right]=\frac{\left[ML{T}^{-2}\right]}{\left[IT\right]}=\left[ML{T}^{-3}{I}^{-1}\right]$ where $\left[I\right]=$ dimension of current.

We know that magnetic force $F=qvB$ or $B=\frac{F}{qv}$

So, $\left[B\right]=\frac{\left[ML{T}^{-2}\right]}{\left[IT\right]\left[L{T}^{-1}\right]}=\left[M{T}^{-2}{I}^{-1}\right]$

Thus, $\frac{\left[E\right]}{\left[B\right]}=\frac{\left[ML{T}^{-3}{I}^{-1}\right]}{M{T}^{-2}{I}^{-1}}=\left[L{T}^{-1}\right]$