🔑:## Step 1: Identify the current distribution and the point of interest.The current is uniformly distributed through a cross-section of the conductor and is directed out of the computer screen. The point of interest, P1, is located at a distance r from the center of the conductor.## Step 2: Determine the relevant geometry for calculating the magnetic field.Given that the conductor has a radius of 'a' and there are two cylindrical cavities of diameter 'a' through its entire length, we need to consider how the current flows around these cavities. However, for the magnetic field at point P1, we primarily consider the current distribution in relation to the distance from the center of the conductor.## Step 3: Apply Ampère's Law to find the magnetic field.Ampère's Law states that the line integral of the magnetic field around a closed loop is proportional to the current enclosed by the loop. For a long, straight wire or a cylindrical conductor, the magnetic field at a distance 'r' from the center can be found using the formula derived from Ampère's Law.## Step 4: Consider the effect of the cavities on the magnetic field calculation.Since the cavities are symmetrically placed and the current is uniform, their presence affects the overall current density but not the basic approach to calculating the magnetic field at a point outside the conductor. The key factor is the total current and the distance from the axis of the conductor.## Step 5: Derive the equation for the magnetic field at point P1.For a point outside a long, cylindrical conductor carrying a current I, the magnetic field (B) at a distance 'r' from the center of the conductor can be calculated using the formula B = μ₀I / (2πr), where μ₀ is the magnetic constant (permeability of free space).The final answer is: boxed{frac{mu_0 I}{2pi r}}

🔑:## Step 1: Calculate the self-energy of the electron as a spherical shellThe self-energy of a charged spherical shell can be calculated using the formula for the energy stored in an electric field, which is given by (U = frac{1}{2} int mathbf{E} cdot mathbf{D} , dV), where (mathbf{E}) is the electric field and (mathbf{D}) is the electric displacement field. For a spherical shell with charge (Q) uniformly distributed over its surface, the electric field inside the shell is zero, and outside the shell, it is (E = frac{Q}{4piepsilon_0r^2}), where (r) is the distance from the center of the shell and (epsilon_0) is the vacuum permittivity.## Step 2: Calculate the electric field everywhere in spaceFor (r < r_0), the electric field is (E = 0) because the charge is distributed on the surface of the shell, and by Gauss's law, the electric field inside a spherical shell is zero. For (r geq r_0), the electric field is (E = frac{Q}{4piepsilon_0r^2}), where (Q) is the total charge of the electron, which is (-e) (considering (e) as the elementary positive charge).## Step 3: Calculate the total energy stored in the fieldThe energy stored in the electric field can be calculated using the formula (U = frac{1}{2} int mathbf{E} cdot mathbf{D} , dV). Since (mathbf{D} = epsilon_0 mathbf{E}), the energy stored becomes (U = frac{epsilon_0}{2} int E^2 , dV). For the region outside the shell ((r geq r_0)), substituting (E = frac{Q}{4piepsilon_0r^2}) into the integral gives (U = frac{epsilon_0}{2} int_{r_0}^{infty} left(frac{Q}{4piepsilon_0r^2}right)^2 4pi r^2 , dr).## Step 4: Evaluate the integral for the energySubstituting the expression for (E) into the integral and simplifying yields (U = frac{1}{2} int_{r_0}^{infty} left(frac{Q}{4piepsilon_0r^2}right)^2 4pi r^2 , dr = frac{Q^2}{8piepsilon_0} int_{r_0}^{infty} frac{1}{r^2} , dr). Evaluating the integral gives (U = frac{Q^2}{8piepsilon_0} left[-frac{1}{r}right]_{r_0}^{infty} = frac{Q^2}{8piepsilon_0r_0}).## Step 5: Take the limit as r0 -> 0As (r_0) approaches 0, the energy (U) approaches infinity because (U = frac{Q^2}{8piepsilon_0r_0}). This implies that the self-energy of a point charge (in this case, an electron) is infinite.The final answer is: boxed{infty}

🔑:## Step 1: Initial ConditionThe fluid in the container is initially accelerating with an acceleration a_1. According to Newton's second law of motion, the force acting on the fluid is given by F = ma_1, where m is the mass of the fluid. This acceleration causes the fluid to exert a pressure on the container walls, with the pressure being highest at the bottom of the container due to the weight of the fluid and the acceleration.## Step 2: Change in AccelerationWhen the acceleration of the container instantly changes to a_2 (where a_2 < a_1), the force acting on the fluid changes to F = ma_2. Since a_2 is less than a_1, the new force F is less than the initial force. This reduction in force results in a decrease in the pressure exerted by the fluid on the container walls.## Step 3: Fluid Motion Relative to the ContainerAs the container's acceleration decreases, the fluid, due to its inertia, tends to continue moving in the original direction of acceleration (a_1) relative to the container. This means the fluid will initially appear to move towards the front of the container (in the direction of a_1) because the container is now accelerating at a slower rate (a_2).## Step 4: Equilibrium and Final StateOver time, the fluid will come to a new equilibrium state where it is again at rest relative to the container, but the distribution of pressure within the fluid will have adjusted to the new acceleration a_2. The pressure gradient within the fluid will be less steep than before because a_2 < a_1, resulting in a lower pressure difference between the bottom and the top of the container compared to the initial state.## Step 5: Physical PrinciplesThis description is based on Newton's laws of motion, particularly the first law (inertia) which explains why the fluid continues to move in the direction of the original acceleration when the container's acceleration changes, and the second law, which relates the force acting on the fluid to its acceleration.The final answer is: boxed{a_2 < a_1}

🔑:## Step 1: Determine the shear rate at the wall of the veinTo calculate the shear stress at the wall of the vein, we first need to determine the shear rate. The shear rate (γ) can be estimated using the formula γ = 4V/D for a Newtonian fluid in a cylindrical pipe, where V is the average velocity and D is the diameter of the pipe. However, since blood is non-Newtonian, this formula provides a rough estimate. Given V = 15 cm/s and D = 2.5 cm, we can calculate the shear rate as γ = 4*15/2.5 = 24 sec^-1.## Step 2: Identify the viscosity equation for bloodThe viscosity of blood, which is non-Newtonian, can be described by various models. A common model for blood viscosity (μ) as a function of shear rate (γ) is the power-law model, μ = μ0 * (γ0 / γ)^n, where μ0 and γ0 are reference viscosity and shear rate, and n is the power-law index. However, without specific values for these parameters, we'll consider a more general approach to understanding how shear stress (τ) relates to shear rate, given by τ = μ * γ.## Step 3: Calculate the shear stress at the wall of the veinFor a non-Newtonian fluid like blood, the shear stress at the wall of a cylindrical pipe can be more complex to calculate directly without a specific viscosity model. However, we can use the general formula τ = μ * γ, where μ is the viscosity at the given shear rate. Without a specific viscosity equation provided for blood, we acknowledge that the viscosity of blood decreases as the shear rate increases, a characteristic of pseudoplastic fluids.## Step 4: Sketch the relationship between shear rate and shear stress for bloodGiven that blood behaves as a pseudoplastic (or shear-thinning) fluid, its viscosity decreases with increasing shear rate. Thus, the relationship between shear rate and shear stress is not linear, as it would be for a Newtonian fluid. At low shear rates, the viscosity is high, and the shear stress increases slowly with shear rate. As the shear rate increases, the viscosity decreases, leading to a more rapid increase in shear stress. However, without a specific equation for blood's viscosity, we cannot precisely plot this relationship but can describe its general shape.## Step 5: Consider body forces and continuityBody forces, such as gravity, can affect fluid flow, but in the context of blood flow in a vein, their impact is typically negligible compared to the forces driving the flow (e.g., pressure differences). Continuity, which states that mass flow rate is constant throughout a pipe (ρ * A * V = constant), is satisfied in a steady, incompressible flow like that in a large vein, assuming no significant changes in density or cross-sectional area.The final answer is: boxed{24}