The Meissner Effect: A fundamental property of a type 1 superconductor is perfec

Question

The Meissner Effect: A fundamental property of a type 1 superconductor is perfect diamagnetism, demonstrated by the Meissner effect illustrated in the photograph of the levitating magnet. The superconducting material has a magnetic field B = 0 everywhere inside it. If a sample of the material is placed intro an externally produced magnetic field or if it is cooled to become superconducting while it is in a magnetic field, electric currents appear on the surface of the sample. These currents have precisely the strength and orientation to make the total magnetic field zero throughout the interior of the sample. The following problem deals with the magnetic force acting on a superconducting sample.

A vertical solenoid with a length of 120 cm and a diameter of 2.5 cm consists of 1400 turns of copper wire carrying a counterclockwise current of 2.0 A. (shown in Fig. P23.47 (a)).

(a) Find the magnetic field in the vacuum inside the solenoid.

(b) Find the energy density of the magnetic field. Note that the unit Jm-3 are the same as the units Nm-2 of pressure.

(c) Now a superconducting bar is inserted inside the solenoid. Its upper end is far outside the solenoid, where the magnetic field is negligible. The lower end of the bar is deep inside the solenoid. Identify the direction required for the current on the curved surface of the bar so that the total magnetic field inside the bar is zero. The field created by the supercurrents is sketched in Fig. 23.47(b), and the total field B? is total sketched in Fig. P23.47 (c).

(d) The field of the solenoid exerts a force on the current in the superconductor. Identify the direction of the force on the bar.

(e) Calculate the magnitude of the force by multiplying the energy density of the solenoid field times the area of the bottom end of the superconducting bar.

Explanation / Answer

Part a:
Magnetic field=
B = mu * N * L / I
N = 1400 / 1.20m = 1166.66 turns per meter
mu o = 1.25 * 10 ^ (-6)
i = 2 A
L = 1.2 m
so,
B = 8.79 * 10 ^ -4

part 2:
As per the question: the unit Jm-3 are the same as the units Nm-2 of pressure.
pressure = Force / area
= ILB / (pi * .025m^2)
=1.07 jm-3

Part 3:
Direction is as per diagram "c"

Part 4:
Force is due UPWARDS

Part 5:
Magnitude of Force = Area * energy desity
= 1.07 * (pi * .025m^2)
=2.1 * 10 ^(-3) N

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