1.

The dimensions of the ratio of magnetic flux (phi) and permeability. (mu) are

Answer»

`[M^(0)L^(1)T^(0)A^(1)]`
`[M^(0)L^(-3)T^(0)A^(1)]`
`[M^(0)L^(1)T^(1)A^(-1)]`
`[M^(0)L^(2)T^(0)A^(1)]`

Solution :MAGNETIC flux = `phi` = BA, B = magnetic filed, A = AREA Permeability = `mu=B/H`.
`phi/mu=(BA)/((B/H))="Area"xx"magnetic intensity"`
[Area] = [A] = [`L^(2)`]
Magnetic intensity = H = nI = `("NUMBER of turns")/("meter")xx"current"`
`[H]=[A^(1)/L]" "["Current"]=[A^(1)]`
`therefore[phi/mu]=[L^(2)xxA^(1)/L]=[LA^(1)]`
`therefore[phi/mu]=[M^(0)L^(1)T^(0)A^(1)]`


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