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Standard XII

Physics

Current Loop Is a Magnet?

Given that ...

Question

Given that $I$ = moment of inertia, $P_{m} =$ magnetic pole momentum and B $=$ magnetic induction, then the dimensional formula for $\frac{I}{P_{m} B}$ is same as that of:

t i m e

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l e n g t h

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{t i m e}^{2}

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{l e n g t h}^{2}

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Solution

The correct option is C ${t i m e}^{2}$
$\frac{I}{P_{m} B} = \frac{M^{1} L^{2}}{I L^{2} M^{1} I^{- 1} T^{- 2}} = T^{2}$ = $t i m e^{2}$

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Similar questions

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Given that I= moment of inertia, Pm= magnetic pole momentum and B= magnetic induction, then the dimensional formula for IPmB is same as that of:

The correct option is C time2IPmB=M1L2IL2M1I−1T−2=T2 = time2

Given that $I$ = moment of inertia, $P_{m} =$ magnetic pole momentum and B $=$ magnetic induction, then the dimensional formula for $\frac{I}{P_{m} B}$ is same as that of:

The correct option is C ${t i m e}^{2}$
$\frac{I}{P_{m} B} = \frac{M^{1} L^{2}}{I L^{2} M^{1} I^{- 1} T^{- 2}} = T^{2}$ = $t i m e^{2}$