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Chapter 15: Problem 13

Curie temperature is the temperature above which (A) a paramagnetic material becomes diamagnetic. (B) a ferromagnetic material becomes diamagnetic. (C) a paramagnetic material becomes ferromagnetic. (D) a ferromagnetic material becomes paramagnetic.

Short Answer

Expert verified
The correct answer is option (D) - A ferromagnetic material becomes paramagnetic above the Curie temperature.

Step by step solution

01

Understanding Curie Temperature

Curie temperature is the critical temperature at which a magnetic material changes its magnetic characteristics. This temperature plays a crucial role in defining the magnetic properties of a material.
02

Option (A)

A paramagnetic material becoming diamagnetic: Paramagnetic materials are those that are attracted to an external magnetic field. Diamagnetic materials are those that repel external magnetic fields. However, the Curie temperature is not associated with the change between paramagnetic and diamagnetic materials. Hence, this option is incorrect.
03

Option (B)

A ferromagnetic material becoming diamagnetic: Ferromagnetic materials are those that have strong attraction towards external magnetic fields and have the property to retain their magnetic properties even after the external magnetic field is removed. Diamagnetic materials repel external magnetic fields. The change between ferromagnetic to diamagnetic is also not associated with the Curie temperature. Hence, this option is incorrect.
04

Option (C)

A paramagnetic material becoming ferromagnetic: The Curie temperature is associated with a change in magnetic properties, but not with the transformation of a paramagnetic material into a ferromagnetic material. Hence, this option is incorrect.
05

Option (D)

A ferromagnetic material becoming paramagnetic: When a ferromagnetic material is heated to its Curie temperature, its magnetic domains become randomized, and the material loses its ferromagnetic properties. Above the Curie temperature, the material is converted into a paramagnetic material, in which it weakly attracts an external magnetic field but does not have the ability to retain its magnetic properties once the external magnetic field is removed. This statement is correct.
06

Final Answer

Based on the analysis above, the correct answer is option (D) - A ferromagnetic material becomes paramagnetic above the Curie temperature.

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Most popular questions from this chapter

A proton with a speed \(u\) along the positive \(x\)-axis at \(y=0\) enters a region of uniform magnetic field \(B=-B_{0} \hat{k}\) which exists to the right of \(y\)-axis. The proton exits from the region after some time with the speed \(v\) at ordinate \(y\), then (A) \(v>u, y<0\) (B) \(v=u, y>0\) (C) \(v>u, y>0\) (D) \(v=u, y<0\)

Current \(i\) is carried in a wire of length \(L\). If the wire is turned into a circular coil, the maximum magnitude of torque in a given magnetic field \(B\) will be (A) \(\frac{L^{2} i B}{2}\) (B) \(\frac{L^{2} i B}{\pi}\) (C) \(\frac{L^{2} i B}{4 \pi}\) (D) \(\frac{L i^{2} B}{4 \pi}\)

A metal disc of radius \(R=6 \mathrm{~cm}\) is mounted on a frictionless axle. The current can flow through the axle out along the disc to a sliding contact of rim of the disc. A uniform magnetic field \(B=2 \mathrm{~T}\) is parallel to the axis of the disc. When the current is \(3 \mathrm{~A}\), the disc rotates with constant angular velocity. The frictional force at the rim between the stationary electrical contact and the rotating rim is \(9 x \times 10^{-2} \mathrm{~N}\). Find the value of \(x\).

A triangular loop of side \(l\) carries a current \(I .\) It is placed in a magnetic field \(B\) such that the plane of the loop is in the direction of \(B\). The torque on the loop is (A) Zero (B) \(I B l^{2}\) (C) \(\frac{\sqrt{3}}{2} I B l^{2}\) (D) \(\frac{\sqrt{3}}{4} I B l^{2}\)

Two short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\), such that their centers coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{M \sqrt{5}}{d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \cdot \frac{3 M}{d^{3}}\)
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