1.

A long solenoid has 1000 turns per meter and carries a current of 1 A. It has a soft iron core of mu_(r) = 1000. The core is heated beyond the Curie temperature, T_(C).

Answer»

The H field in the solenoid is (nearly) unchanged but the B field decreases drastically.
The H and B FIELDS in the solenoid are nearly unchanged.
The MAGNETISATION in the core reverses direction.
The magnetisation in the core diminishes by a factor of about `10^8`.

Solution :Magnetic field intensity of solenoid,
`H= nI""` (Where `n=` number of turns per meter of a solenoid)
`therefore H= 1000 xx 1 = 1000` Am
Hence, H is a constant, so it is nearly unchanged.
`But, B= mu n I`
`= mu_(r) mu_(0) nI`
`= (mu_(0) nI) mu_(r) = (K=` constant `)mu_(r)`
`therefore B prop mu_r`
Hence, if `mu_r` changes, B also changes.
Now for magnetisation in the core, when temperature of the iron core of a solenoid is RAISED beyond Curie temperature, then it behaves as a paramagnetic material where,
`(chi_(x) )_("Fero") ~~ 10^(3)`
`(chi_(m) )"(Para") ~~ 10^(-5) `
`((chi_(m) )_("Fero"))/( (chi_(m))_("Para") = (10^(3) )/( 10^(-5) ) = 10^(8)`
Hence, magnetization of core decreases by `10^(8)` 7) times hence option (D) is correct.


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