Question 64: Units , Dimensions and Error Analysis

Question

If $B$ is magnetic field and $\mu_0$ is permeability of free space, then the dimensions of $\left(B / \mu_0\right)$ is

Accepted Answer

To determine the dimensions of $\left(\frac{B}{\mu_0}ight)$, we start with the formula for the magnetic field $B$ inside a solenoid:
$ B = \mu_0 n I $
where $n = \frac{N}{L}$ is the number of turns per unit length, $I$ is the current, and $\mu_0$ is the permeability of free space.
Rearranging the formula gives:
$ \frac{B}{\mu_0} = nI = \frac{N}{L}I $
From the above, the dimensions of $\left(\frac{B}{\mu_0}ight)$ are:
$ \left[\frac{B}{\mu_0}ight] = \left[L^{-1} Aight] $
Thus, the dimensional formula of $\left(\frac{B}{\mu_0}ight)$ corresponds to $L^{-1} 	ext{A}$.

Answer

$\mathrm{MT}^{-2} \mathrm{~A}^{-1}$

Answer

$\mathrm{L}^{-1} \mathrm{~A}$

Answer

$\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~A}^{-1}$

Answer

$\mathrm{LT}^{-2} \mathrm{~A}^{-1}$

Units , Dimensions and Error Analysis

If B is magnetic field and \mu_0 is permeability of free space, then the dimensions of \left(B / \mu_0\right) is

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