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At temperatures near absolute zero, \(B_{\mathrm{c}}\) approaches \(0.142 \mathrm{~T}\) for vanadium, a type-I superconductor. The normal phase of vanadium has a magnetic susceptibility close to zero. Consider a long, thin vanadium cylinder with its axis parallel to an external magnetic field \(\overrightarrow{\boldsymbol{B}}_{0}\) in the \(+x\) -direction. At points far from the ends of the cylinder, by symmetry, all the magnetic vectors are parallel to the \(x\) -axis. At temperatures near absolute zero, what are the resultant magnetic field \(\vec{B}\) and the magnetization \(\vec{M}\) inside and outside the cylinder (far from the ends) for (a) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.130 \mathrm{~T}) \hat{\boldsymbol{\imath}}\) and (b) \(\overrightarrow{\boldsymbol{B}}_{0}=(0.260 \mathrm{~T}) \hat{\imath} ?\)

Short Answer

Expert verified
For \(B_{0} = 0.130 T\), the resultant magnetic field inside the superconducting cylinder is \(B = 0 T\) and the induced magnetization \(M = -B_{0}/\mu_{0}\). For \(B_{0} = 0.260 T\), the resultant magnetic field is \(B = 0.260 T\) and the induced magnetization \(M = 0 A/m\)

Step by step solution

01

Determining the state of the superconductor under the applied field

Determine whether the superconductivity in vanadium will remain or will be lost under the applied external magnetic field \(B_{0}\). Compare the value of the magnetic field \(B_{0}\) with the critical magnetic field \(B_{c}\). If \(B_{0} < B_{c}\), vanadium will remain superconducting, and if \(B_{0} > B_{c}\), superconductivity will be lost.
02

Calculating the magnetic field and magnetization for \(B_{0} = 0.130 T\)

When \(B_{0} = 0.130 T\), which is less than \(B_{c}\) of vanadium, superconductivity will remain. Hence, the magnetic field inside the superconductor will be zero, i.e., \(B=0 T\). Since superconductors repel all magnetic field lines when below their critical fields, the magnetization induced in the superconductor will be in the opposite direction to the applied field. This can be calculated as \(M = -B_{0}/\mu_{0}\), where \(\mu_{0}\) is the permeability of free space.
03

Calculating the magnetic field and magnetization for \(B_{0} = 0.260 T\)

When \(B_{0} = 0.260 T\), which is more than \(B_{c}\) of vanadium, superconductivity will be lost. Therefore, the material will behave as a normal conductor. The magnetic field inside the conductor will be equal to the external magnetic field \(B_{0}\), i.e., \(B=B_{0} = 0.260 T\). The magnetization of a normal material is expected to be close to zero, i.e., \(M=0 A/m\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Susceptibility
Magnetic susceptibility is a measure of how much a material becomes magnetized when exposed to an external magnetic field. This property of materials is particularly interesting when studying superconductors. In the context of our vanadium example, at temperatures near absolute zero, its magnetic susceptibility is close to zero in the normal phase. This implies that when not superconducting, vanadium doesn't magnetize significantly in response to an external magnetic field.

However, when vanadium enters the superconducting state, magnetic susceptibility plays a vital role. In a superconductor, magnetic susceptibility becomes effectively negative because the material exhibits perfect diamagnetism – it repels the applied magnetic field completely. This is also known as the Meissner effect. Thus, when a superconductor like vanadium is below its critical temperature and within its critical magnetic field, it will experience a perfect diamagnetic response, leading to a resulting magnetic field inside the material of zero.
Critical Magnetic Field
The critical magnetic field (\(B_c\)) of a superconductor is a fundamental concept that defines the threshold above which the superconducting state is destroyed. When the externally applied magnetic field exceeds this critical value, the material transitions from the superconducting to the normal conducting state. For vanadium, a Type-I superconductor, the critical magnetic field near absolute zero is given as 0.142 Tesla.

In our exercise, the critical magnetic field serves as a benchmark to determine the state of vanadium under different external magnetic fields. When the applied field (\(B_0\)) is less than the critical field (\(B_0 < B_c\)), the vanadium remains in the superconducting state, exhibiting zero internal magnetic field. But if the applied field is greater (\(B_0 > B_c\)), the superconductivity is lost, and the internal magnetic field will equal the external field.
Magnetization
Magnetization (\( \textbf{M} \)) of a material is the vector field that expresses the density of permanent or induced magnetic dipole moments in the material. In superconductors like vanadium, magnetization becomes quite interesting. When the external magnetic field is below the critical field (\(B_0 < B_c\)), the superconductor will exhibit complete expulsion of the magnetic field lines, a state of perfect diamagnetism. As we infer from the exercise, the magnetization is then in the opposite direction to the applied magnetic field and is determined by \(M = -\frac{B_0}{\text{µ}_0}\), where \(\mu_0\) is the magnetic permeability of free space.

In the case where the applied field surpasses the critical value (\(B_0 > B_c\)), vanadium will lose its superconducting properties, and the magnetization is expected to be close to zero due to its low magnetic susceptibility in the normal state. Here, the internal magnetic field is just the applied one, and no significant magnetization is observed.
Type-I Superconductor
Type-I superconductors are materials that have a single critical magnetic field, distinguishing them from Type-II superconductors, which have two critical fields. Vanadium, the material in question, is a Type-I superconductor. These materials are characterized by complete magnetic flux expulsion (the Meissner effect) below this critical magnetic field, as opposed to the partial expulsion seen in Type-II superconductors.

Type-I superconductors also tend to have a low critical temperature and critical magnetic field. They do not exhibit mixed states or vortices, as Type-II superconductors do, which allows them to maintain zero electrical resistance and expel magnetic fields until the applied field reaches the critical threshold. Once the critical field is exceeded, the superconducting state collapses abruptly, and the material reverts to normal conductivity, allowing the magnetic field to penetrate it completely.

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Most popular questions from this chapter

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