A coil of wire \(0.5 \mathrm{~m}\) long and having 20 turns carries a current of \(1.0 \mathrm{~A}\). (a) Compute the flux density if the coil is within a vacuum. (b) A bar of an iron-silicon alloy, the \(B-H\) behavior for which is shown in Figure \(20.29\), is positioned within the coil. What is the flux density within this bar? (c) Suppose that a bar of molybdenum is now situated within the coil. What current must be used to produce the same \(B\) field in the Mo as was produced in the iron-silicon alloy (part b) using \(1.0 \mathrm{~A}\) ?

To compute the flux density within a vacuum, we need to first find the magnetic field produced by a coil with a certain current flowing through it. We can use Ampere's Law for a solenoid (coil) here: \[\vec{B}=\mu_{0} * n * \vec{I}\] Where, \(\vec{B}\) is the magnetic field, \(\mu_{0}\) is the permeability of vacuum (\(4\pi\times10^{-7} Tm/A\)), \(n\) is the number of turns per unit length, \(I\) is the current through the coil. Given: Length of wire coil = \(0.5 m\), Number of turns = \(20\), Current = \(1.0 A\). We can calculate the number of turns per unit length: \[n = \frac{Number~of~turns}{Length} = \frac{20}{0.5} = 40~ turns/m\] Now, we can compute the magnetic field \(B\) using the formula: \[B=\mu_{0} * n * I=(4\pi\times10^{-7}) * (40) * (1.0) = 5.027 \times 10^{-5} T\] So, the flux density \(B\) is \(5.027 \times 10^{-5} T\) in vacuum.

We are given the \(B-H\) behavior curve for the iron-silicon alloy. The magnetic field strength \(H\) is related to the current \(I\) by the relationship \(H=nI\). From the previous part, we found the number of turns per length \(n = 40\) and the given current is \(I = 1\) A. Calculate \(H\): \[H = nI = (40)(1.0) = 40 A/m\] Now, to find the flux density within the iron-silicon alloy, we can read the value of \(B\) from the \(B-H\) curve given in the problem when \(H = 40 A/m\). There is no numerical value available here, so we'll refer to the value of \(B\) as \(B_{Fe-Si}\), which corresponds to \(H=40 A/m\) on the iron-silicon \(B-H\) curve.

We are given that we need to find the current required to produce the same \(B\) field in molybdenum as was produced in the iron-silicon alloy for the 1 A current in part (b). We know that for molybdenum, \(B=\mu_{Mo} * H_{Mo}\). We can rearrange the formula to find the value of \(H_{Mo}\): \[H_{Mo} = \frac{B_{Fe-Si}}{\mu_{Mo}}\] Now, we'll get the required value of current using the previously known relationship between \(H\) and \(I\): \[H_{Mo} = nI_{Mo}\] Solving for \(I_{Mo}\): \[I_{Mo} = \frac{H_{Mo}}{n} = \frac{B_{Fe-Si}}{n\mu_{Mo}}\] We are not given the value of \(\mu_{Mo}\) for molybdenum in the problem, so we cannot get a numerical value for the current \(I_{Mo}\). Nevertheless, the formula above illustrates how to calculate the current required to produce the same \(B\) field in molybdenum as the one achieved in the iron-silicon alloy.