Question

At temperatures near absolute zero, Bc approaches 0.142 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 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.

Part B

What is the direction of the resultant magnetic field B? inside the cylinder for this case?

What is the direction of the resultant magnetic field inside the cylinder for this case?

in the +x-direction
in the ?x-direction
perpendicular to the x-axis
the field is zero
Part C

What is the magnitude of the resultant magnetic field B? outside the cylinder (far from the ends) for this case?

Part E

What is the magnitude of the magnetization M? inside the cylinder for this case?

Part G

What is the magnitude of the magnetization M? outside (far from the ends) the cylinder for this case?

Answer 1

b) field is zero, c) the magnetic field does not change in intensity or direction

e) M = -H = Bo /μ₀ , g) M = 0

Step-by-step explanation:

Part b

superconductors are formed by so-called Coper pairs that are electrons linked through a distortion in the network, this creates that they must be treated as an entity so we have an even number of charge carriers and the material must behave with diamagnetic , Meissner effect, consequently the magnetic field inside its superconductor is zero

the correct answer is Zero

Part c

outside the superconducting cylinder the magnetic field does not change in intensity or direction

Part E

Magnetization is defined by the equation

B = μ₀ (H + M)

with field B it is zero inside the superconductors

M = -H = Bo /μ₀

where Bo is the magnetic induction in the normal state

Part g

As outside the cylinder there is no material zero magnetization

M = 0