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An iron rod is placed parallel to magnet...

An iron rod is placed parallel to magnetic field of intensity `2000 Am^-1`. The magnetic flux through the rod is `6xx 10^-1` Wb and its cross-sectional area is `3 cm^2`. The magnetic permeability of the rod in Wb `A^-1 m^-1` is

A

`10^(-1)`

B

`10^(-2)`

C

`10^(-3)`

D

`10^(-4)`

Text Solution

Verified by Experts

The correct Answer is:
C
`H=2000A//m,phi=6xx10^(-4)k//b,A=3cm^(2)mu=?`
`mu=(B)/(H)=(phi)/(AH)(":'B=(phi)/(A))`
`:.mu=(6xx10^(-4))/(3xx10^(-4)xx2000)=10^(-3)` Wb/Am.
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Knowledge Check

  • A iron rod is placed parallel to magnetic field of intensity 2000 A/m The magnetic flux through the rod is 6xx10^(-4) Wb its cross sectional area is 3 cm^(2) The magnetic permeability of the rod in Wb/A -m is

    A
    `10^(-1)`
    B
    `10^(-2)`
    C
    `10^(-3)`
    D
    `10^(-4)`

  • A magnetising field fo 1000 A/m produces a magnetics flux of 2.4 xx10^(-5) Wb in an iron bar of cross sectional area 0.3 cm^(2) what is the magnetic permeability of the iron bar ?

    A
    weber metre `"ampere"^(-1)`
    B
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    D
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    B
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    C
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    D
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