Question
Derive dimension and SI unit of magnetic flux.
✓
Answer
Magnetic Flux : Magnetic flux linked with a surface of area A placed in a magnetic field B is given by $\Phi_{B} = \vec{B}.\vec{A}$ i.e. $\Phi_{ B }= B \cdot A \cos \theta$, where $\theta=$ angle between $\overrightarrow{ B }$ and $\overrightarrow{ A }$.
Dimension of $\Phi_{B}=[B][A]$
(because $\cos\theta$ is dimensionless)
$\Rightarrow \quad\left[\Phi_{ B }\right]=\left[ MT ^{-2} A^{-1}\right]\left[ L ^2\right]$
$=[ML^{2}T^{-2}A^{-1}]$
S.I. Unit of Magnetic Flux :
$\Phi_{B}=BA\cos\theta$
⇒ S.I. Unit of $\Phi_{B}=S.I.$ Unit of $B \times S.I.$ Unit of A
$[\because \cos \theta$ has no unit $]$
$=(\frac{N}{Am}) \times m^{2}=NmA^{-1}$
Since $\quad Nm=J$
$\therefore \quad$ S.I. unit of $\Phi_B=J A^{-1}$
Although the S.I. unit of magnetic flux used in practice is weber denoted by Wb.
Dimension of $\Phi_{B}=[B][A]$
(because $\cos\theta$ is dimensionless)
$\Rightarrow \quad\left[\Phi_{ B }\right]=\left[ MT ^{-2} A^{-1}\right]\left[ L ^2\right]$
$=[ML^{2}T^{-2}A^{-1}]$
S.I. Unit of Magnetic Flux :
$\Phi_{B}=BA\cos\theta$
⇒ S.I. Unit of $\Phi_{B}=S.I.$ Unit of $B \times S.I.$ Unit of A
$[\because \cos \theta$ has no unit $]$
$=(\frac{N}{Am}) \times m^{2}=NmA^{-1}$
Since $\quad Nm=J$
$\therefore \quad$ S.I. unit of $\Phi_B=J A^{-1}$
Although the S.I. unit of magnetic flux used in practice is weber denoted by Wb.
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