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The dimensions of permeability of free s...

The dimensions of permeability of free space can be given by

A

`[MLT^(-2)A^(-2)]`

B

`[MLA^(-2)]`

C

`[ML^(-3)T^(2)A^(2)]`

D

`[MLA^(-1)]`

Text Solution

Verified by Experts

The correct Answer is:
A
Force per unit length between two parallel current carrying conductor
`F=(mu_(0))/(4pi)(2I_(1)I_(2))/d N//m`
`N/m=mu_(0)/(4pi)(Amp^(2))/m implies mu_(0)=N/Amp^(2)`
`[mu_(0)]=[MLT^(-2)A^(-2)]`
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The magnetic permeability of free space is …………………… .

Dimension of magnetic permeability is :

Knowledge Check

  • Dimensions of permeability are

    A
    `A^(-2) M^(1) L^(1) T^(-2)`
    B
    `MLT^(-2)`
    C
    `ML^(0) T^(-1)`
    D
    `A^(-1) MLT^(2)`

  • The dimensional formula for permeability of free space, mu_(0) is

    A
    `[MLT^(-2)A^(-2)]`
    B
    `[ML^(-1)T^(2)A^(-2)]`
    C
    `[ML^(-1)T^(-2)A^(2)]`
    D
    `[MLT^(-2)A^(-1)]`

  • Dimensions of magnetic permeability is :

    A
    `MLT^(2)A^(-2)`
    B
    `ML^(-1)T^(-2)A^(-2)`
    C
    `ML^(-2)T^(-2)A^(2)`
    D
    `MLT^(-2)A^(-2)`

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    A solenoid of 1000 turns per metre has a core with relative permeability 500. Insulated windings of the solenoid carry an electric current jof 5A. The magnetic flux density produced by the solenoid is : ( permeability of free space = 4 pi xx 10^(-7) H //m )

    When one computes the square root of the ratio of the permeability of free space to the permitivity of free space, one obtains sqrt(mu_(0)/epsilon_(0))=337 . What is the unit of this quantity?

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