a) A loop of wire \(5.00 \mathrm{~cm}\) in diameter is carrying a current of \(2.00 \mathrm{~A}\). What is the energy density of the magnetic field at its center? b) What current has to flow in a straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire?

Using Ampère's law, we can find the magnetic field at the center of the loop. The formula for magnetic field at the center of a loop is: \(B = \frac{\mu_0 I}{2R}\) where \(B\) is the magnetic field, \(\mu_0\) is the permeability of free space \((4\pi \times 10^{-7} \mathrm{Tm/A})\), \(I\) is the current in amperes, and \(R\) the radius in meters. We are given \(I = 2.00 \mathrm{A}\) and the diameter \(D = 5.00 \mathrm{cm}\), which means the radius \(R = 0.025 \mathrm{m}\). Now, we can plug the values into the formula: \(B = \frac{4\pi \times 10^{-7} \mathrm{Tm/A} \times 2 \mathrm{A}}{2 \times 0.025 \mathrm{m}}\) \(B = 2.00 \times 10^{-5} \mathrm{T}\) Now, we have to find the energy density \(u\) of the magnetic field, which can be calculated using the formula: \(u = \frac{1}{2\mu_0} B^2\) Plug in the values of \(\mu_0\) and \(B\): \(u = \frac{1}{2 \times 4\pi \times 10^{-7} \mathrm{Tm/A}} \times (2.00 \times 10^{-5} \mathrm{T})^2\) \(u = 1.60 \times 10^{-4} \mathrm{J/m^3}\) Thus, the energy density of the magnetic field at the center of the loop is \(1.60 \times 10^{-4} \mathrm{J/m^3}\).

We are given that the distance from the straight wire to the point where we want the same energy density is \(4.00 \mathrm{cm}\) or \(d= 0.040 \mathrm{m}\). The magnetic field created by a straight wire can be calculated using the Biot–Savart law: \(B = \frac{\mu_0 I}{2\pi d}\) We will then use the formula for the energy density of the magnetic field again: \(u = \frac{1}{2\mu_0} B^2\) Now our target is to find the current \(I\) that will produce the same energy density \(u\) at a distance \(d\). Let's substitute the magnetic field of a straight wire (\(B = \frac{\mu_0 I}{2\pi d}\)) into the energy density formula and solve for \(I\): \(u = \frac{1}{2\mu_0} \left(\frac{\mu_0 I}{2\pi d}\right)^2\) Now plug in the known values of \(u\), \(d\), and \(\mu_0\) and solve for \(I\): \(1.60 \times 10^{-4} \mathrm{J/m^3} = \frac{1}{2 \times 4\pi \times 10^{-7} \mathrm{Tm/A}} \times \left(\frac{4\pi \times 10^{-7} \mathrm{Tm/A} \times I}{2\pi \times 0.040 \mathrm{m}}\right)^2\) After solving, we find: \(I = 2.83 \mathrm{A}\) Therefore, the current required to flow in the straight wire to produce the same energy density at a point \(4.00 \mathrm{~cm}\) from the wire is \(2.83 \mathrm{A}\).