1.

The dimension of (B^2)/( 2mu_0) , where B is magnetic field and mu_0 is the magnetic permeability of vacuum is

Answer»

`M^(1) L^(-1) T^(-2)`
`M^(1) L^(2) T^(-2)`
`M^(1) L^(-1) T^(2)`
`M^(1) L^(-2) T^(-1)`

Solution :Energy density DUE to MAGNETIC field `= (B^2)/( 2mu_0)`
`("Force" xx" displacement")/("VOLUME") = (B^2)/( 2mu_0) ""[because "Energy" = "Force" xx "displacement" ]`
`therefore [ (B^2) /( 2mu_0) ] = ([F][d])/([V]) ""[because 2mu_(0) " DIMENSIONLESS" ]`
`(M^(1) L^(1) T^(-2) xx L^(1) )/( L^(3) )`
`= M^(1) L^(-1) T^(-2) `


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