Question

The dimensional formula for magnetic moment of a magnet is :

• A. $\left[M^0{L}^2{T}^0{A}^1\right]$
• B. $\left[M^0{L}^2{T}^0{A}^{-1}\right]$
• C. $\left[M^0{L}^{-2}{T}^0{A}^{-1}\right]$
• D. $\left[M^1{L}^{-2}{T}^0{A}^{-1}\right]$

Answer 1

Answer: (A)

$\left[M^0{L}^2{T}^0{A}^1\right]$

We know that magnetic moment for a solenoid is NIS, where N is number of loops, I is current and S is the vector area.
Thus, dimensions of $\mu$ are $\left[A{L}^2\right]=\left[M^0{L}^2{T}^{0}A\right]$