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Let [epsilon(0)] denote the dimensional ...

Let `[epsilon_(0)]` denote the dimensional formula of the permittivity of the vacuum and `[mu_(0)]` that of the permeability of the vacuum. If M= mass, L=length, T= time and I= electric current

A

`[epsilon_(0)]=[M^(-1)L^(-3)T^(2)I]`

B

`[epsilon_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`

C

`[mu_(0)]=[MLT^(-2)I^(-2)]`

D

`[mu_(0)]=[ML^(2)T^(-1)I]`

Text Solution

Verified by Experts

The correct Answer is:
B, C
`F=(1)/(4pi in_(0))(q_(1)q_(2))/r^(2) rArr MLT^(-2)=[1/in_(0)](A^(2)T^(2))/L^(2)`
`rArr [1/in_(0)]=M^(1)L^(3)A^(-2)T^(-4) rArr [in_(0)]=M^(-1)L^(-3)A^(+2)T^(4)`
`F/L=mu_(0)/(4pi)(i_(1)i_(2))/(r)`
`[ML^(-2)]=[(mu_(0)A^(2))/L]rArr [mu_(0)]=MLA^(-2)T^(-2)`
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Knowledge Check

  • Let [epsi_(0)] denote the dimensional formula of the permittivity of the vacuum and [mu_(0)] that of the permeability of the vacuum. If M = mass, L = length, T = time and I = electric current :

    A
    `[epsi_(0)]=[M^(-1)L^(-3)T^(2)I]`
    B
    `[epsi_(0)]=[M^(-1)L^(-3)T^(4)I^(2)]`
    C
    `[mu_(0)]=[MLT^(-2)I^(-2)]`
    D
    `[mu_(0)]=[ML^(2)T^(-1)I]`

  • Let [in_(0)] denote the dimensional formula of the permittivity of the vacuum and [mu_(0)] that of the permability of the vacuum.If M="mass",L="length",T-"time" and I="electric current"

    A
    `[in_(0)]=M^(-1)L^(-3)T^(2)I`
    B
    `[in_(0)]=M^(-1)L^(-3)T^(4)I^(2)`
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    A
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    B
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    C
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