Q160 / 180 · KCET 2018 · Physics
The dimensions of the ratio of magnetic flux \( (\phi) \) and permeability \( (\mu) \) are
- Correct answerA\( \left[M^{0} L^{1} T^{0} A^{1}\right] \)
- B\( \left[M^{0} L^{-3} T^{0} A^{1}\right] \)
- C\( \left[M^{0} L^{1} T^{1} A^{(-1)}\right] \)
- D\( \left[M^{0} L^{2} T^{0} A^{1}\right] \)
Solution
Magnetic flux, $\phi=B A \rightarrow(1)$
where $B$ is magnetic field; $A$ is area.
Also, $B=\mu H \rightarrow(2)$
where $\mu$ is permeability; $H$ is magnetic field intensity
Substituting Eq. (2) in Eq. (1), we get
$\phi=(\mu H) A$
or $\frac{\phi}{\mu}=H A$
Now, Area A has dimensions $L^{2}$
And magnetic field intensity $H=\frac{\text { Number of turns } \times \text { Current }}{\text { Metre }}$
$\Rightarrow H$ has dimensions $L^{-1} A .$
Therefore, $\frac{\phi}{\mu}=L^{-1} A \times L^{2}=L A$
Hence, the dimensions of ratio of magnetic flux and permeability $\mu$ are $\left[M^{0} L^{1} T^{0} A^{1}\right]$
Hence, the correct answer is option: 'A'.