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A magnetising field of 2xx10^(3)Am^(-1) ...

A magnetising field of `2xx10^(3)Am^(-1)` produces a magnetic flux density of `8piT` in an iron rod. The relative permeability of the rod will be

A

`10^(2)`

B

`10^(0)`

C

`10^(3)`

D

`10^(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
`mu_(r)=(mu)/(mu_(0))=(B)/(mu_(0)H)`
`=(8pi)/((4pixx10^(-7))(2xx10^(3)))`
`mu_(r)=10^(4)`
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