The dimensions of self-inductance L are :

[M L^{2} T^{- 2} A^{- 2}]

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

[M L^{2} T^{- 1} A^{- 2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[M L^{2} T^{- 1} A^{- 1}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

[M L^{- 2} T^{- 2} A^{- 2}]

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $[M L^{2} T^{- 2} A^{- 2}]$
Self-inductance L $= \frac{e}{Δ i / Δ t} = \frac{e Δ t}{Δ i}$
$∴$ Unit of L $= \frac{V o l t - s e c o n d}{A m p e r e}$
$= \frac{(J o u l e / C o u l o m b) s e c o n d}{A m p e r e}$
$= \frac{N e w t o n - m e t r e - s e c o n d}{C o u l o m b - a m p e r e}$
$\Rightarrow$ Unit of L $= \frac{N e w t o n - m e t e r e}{A m p e r e^{2}}$
$∴ [L] = \frac{[M L T^{2}] [L]}{[A^{2}]} = [M L^{2} T^{- 2} A^{- 2}]$

Suggest Corrections

Similar questions

[M L^{2} T^{- 2}]

X

= \sqrt{(\frac{X}{density})}

X

L

L = r p sin θ

r

P

Join BYJU'S Learning Program