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An iron rod of volume 10^(-4)m^(3) and r...

An iron rod of volume `10^(-4)m^(3)` and relative permeability 1000 is placed inside a long solenoid wound with `5 "turns"//"cm"`. If a current of `0.5 A` is passed through the solenoid, then the magnetic moment of the rod is

A

`20Am^(2)`

B

`25Am^(2)`

C

`30Am^(2)`

D

`35Am^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
We know that, `B=mu_(0)H+mu_(0)I`
`therefore I=(B-mu_(0)H)/(mu_(0))=(muH-mu_(0)H)/(mu_(0))=((mu)/(mu_(0))-1)H`
For a solenoid having n turns length and current I.
`" " H=I`
`I=(mu_(r)-I)nl'=(1000-1)500xx0.5`
` " " =2.5xx10^(5)Am^(-1)`
`therefore "Magnetic moment", M=IV=2.5xx10^(5)xx10^(-4)=25A-m^(2)`
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