Ncert Exercises & Solutions

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

(a) The H field in the solenoid is (nearly) unchanged but the B field decreases drastically.
(b) The H and B fields in the solenoid are nearly unchanged.
(c) The magnetization in the core reverses the direction.
(d) The magnetization in the core diminishes by a factor of about $10^{8}$ .

1. (a, c)
2. (a, d)
3. (b, c)
4. (c, d)

(2)

Hint: The magnetisation field intensity does not depend on the material of the core but the magnetisation depends on the material of the core.

Step 1: The magnetic field intensity;

\vec{H} = nI

Magnetisation,

\vec{B} = μ_{o} μ_{r} nI

So the magnetic field intensity remains unchanged and the magnetisation decreases.

Step 2: When the temperature of the iron core of solenoid (which is ferromagnetic materials) is raised beyond curie temperature, then soft iron core behaves as paramagnetic material.

{(χ_{m})}_{ferro} = 10^{5}

{(χ_{m})}_{para} = 10^{- 3}

So, \frac{{(χ_{m})}_{ferro}}{{(χ_{m})}_{para}} = \frac{10^{5}}{10^{- 3}} = 10^{8}

So the magnetization in the core diminishes by a factor of about

10^{8}

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