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Advance Physics

This page explains concepts at college level in the field of quantum mechanics, theory of relativity, superconductivty, Lagrangian method of equation of motion, Solid State Physics, Wave Particle Duality, Diffraction.

- The resistivity of a superconductor is practically zero but only below certain temperature known as Critical Temperature, T
_{C}. - When superconducting materials are cooled below their critical temperatures in the presence of a magnetic field, the magnetic flux is expelled from the interior of the superconductor.
- Similarly, superconducting materials lose their superconducting behavior above a certain critical magnetic field, B
_{C}which itself is temperature-dependent that is B_{C}= B_{C}(T).

- The value of the critical field limits the maximum [critical] current I
_{C}that can be sustained in superconducting materials. - Critical current in a superconducting wire of radius 'a' is given by I
_{C}= 2πB_{C}a/μ_{0}. This is the maximum current the wire can carry at any given temperature and still maintain superconducting features.

A rod of length L is inclined at angle θ with x-axis [in the moving reference frame] at velocity 0.6c where c is the speed of light in vacuum. Calculate the angle observed in a stationary reference frame.

tanθ = L_{Y} / L_{X}.

Since rod the moving along X-axis, the length perpendicular to the direction of motion will remain constant as measured my stationary onserver. Thus: L_{Y}' = L_{Y}.
The contraction in horizontal (X-) component of the length of the rod is given by

The red line shows rod as seen in moving reference frame and the blue line as observed in stationary reference frame. Thus, angle of inclination in stationary reference frame is given by:

tanθ' = L_{Y} / L_{X}' = 1.25 * L_{Y} / L_{X} = 1.25tanθ

A metal behaves as superconductive at 10 [K] and the critical magnetic field at 0 [K] is B_{0} [A/m]. Calculate the magnetic field at 20 [K].

The critical temperature is T_{C} = 10 [K]. The relative between magnetic field and temperature is given by expression:

Thus:
B_{C}(20 K) = B_{0} * (1 - 1/4) = 0.75B_{0}. Note that the magnetic field is specified in terms of [A/m] instead of standard unit of [Tesla]. Here, the specified value is actuall B/μ_{0} where μ_{0} is the permeability of vacuum having units of [T-m/A].

The ground monitoring team of a spaceship moving at constant speed noted the time difference as per his stop-watch as 9 [s] when the location is shown to be 1.44E10^{9} [m]. Calculate the proper time interval

Note that the definition of proper time interval is "time interval measured in a reference frame where two events occcur at same place". Hence, a person monitoring time in space-ship will measure time at same place and will be the proper time interval, Δt. Thus, the time interval measured by the ground staff is improper time interval, Δt'. We have

v = 1.44x10^{9} [m] / 9 [s] = 1.8x10^{8} [m/s], Δt' = 9 [s].

Δt = Δt' * 0.8 = 7.2 [s].

Note that a moving clock runs slower than a stationary clock and hence "proper time interval" ≤ "improper time interval".

Calculate the relativistic speed of a particle of rest mass m_{0} after time t when a constant force F acts on it. The particle stats with velocity v_{0} at t = 0.

From Newtons's law: dp/dt = F. Here, the momentum p = m(v) * v where m(v) is mass of the particle at speed v.
Thus:

Calculate the number of bound states for energy (potential) level E = 11h^{2}/[8mL^{2}] where L is the length of the box and h is the Planck's constant.

The potential well for a particle in a box is shown below

The energy level at quantum state 'n' is given by

Thus, the equation E_{n} < E = 11h^{2}/[8mL^{2}] is satisfied for n = 1, 2 and 3. Hence, number of bound states is 3.

*Note* that the energy in ground state E_{1} > 0, whereas the classical physics assigns a value zero to minimum energy level. This excess energy of the ground state (with respect to the classical minimum) is known as zero point energy. Thus, the kinetic energy hence the momentum of a bound particle cannot be reduced to zero. The minimum value of momentum is E_{1} = p^{2}/2m.

A neutron is moving at velocity 9*10^{7} [m/s]. What is the minimum uncertainty in position of this particle.

The **maximum** uncertainty in position is size of the box, that is Δx = L. Thus, Δx.Δp ≈ h. Similary, **minimum** uncertainty in position will correspond to maximum uncertainty in momentum.

Momentum of the neutron particle is given by

p = m_{n} * v = 1.674929 x 10^{-27} [kg] * 9 x 10^{7} [m/s]

p = 5.165 x 10^{-19} [kg-m/s] = maximum uncertainty in momentum.

Thus: minimum uncertainty in position, Δx_{MIN} = h / Δp_{MAX}

Δx_{MIN} = 6.626 x 10^{-34} [kg-m^{2}/s] / 5.165 x 10^{-19} [kg-m/s] = 1.283 x 10^{-15} [m].

A particle is located in a box of length L and is assumed to be in ground state. Determine the probability of finding the particle in the region 0.25L to 0.75 L.

The probability distribution function (also known as wave function) is given by

The ground state, 'n' = 1 and the square of the wavefunction is related to the probability of finding the particle in a specific position for a given energy level. Thus, cumulative probability between 0.25 to 0.75L is given by

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