At Los Alamos National Laboratories, one means of producing very large magnetic fields is known as the EPFCG (explosively-pumped flux compression generator), which is used to study the effects of a high-power electromagnetic pulse (EMP) in electronic warfare. Explosives are packed and detonated in the space between a solenoid and a small copper cylinder coaxial with and inside the solenoid, as shown in the figure. The explosion occurs in a very short time and collapses the cylinder rapidly. This rapid change creates inductive currents that keep the magnetic flux constant while the cylinder's radius shrinks by a factor of \(r_{\mathrm{i}} / r_{\mathrm{f}}\). Estimate the magnetic field produced, assuming that the radius is compressed by a factor of 14 and the initial magnitude of the magnetic field, \(B_{i}\), is \(1.0 \mathrm{~T}\).

Step by step solution

01

Expression for Magnetic Flux

To begin solving this problem, we will first write the expression for magnetic flux inside a solenoid. Magnetic flux is given by: \(\Phi = nAB\) where \(n\) is the number of loops per unit length of the solenoid, \(A\) is the area of the solenoid's cross-section where the field is present, and \(B\) is the magnetic field inside the solenoid.

02

Use Flux Conservation

According to the problem, the magnetic flux should remain constant throughout the process. Hence, we can write: \(\Phi_i = \Phi_f\) where \(\Phi_i\) is the initial magnetic flux, and \(\Phi_f\) is the final magnetic flux.

03

Plug in Initial and Final Flux Expressions

Now we can write the expressions for the initial and final magnetic fluxes using the formula from step 1: \(n_i A_i B_i = n_f A_f B_f\) Since the number of loops per unit length remains unchanged, \(n_i = n_f = n\). Also, the initial magnetic field is given as \(B_i = 1.0~T\). We now have: \(nA_i (1.0~\mathrm{T}) = n A_f B_f\)

04

Relate Initial and Final Areas

We are given that the radius is compressed by a factor of 14 (i.e., \(\frac{r_i}{r_f} = 14\)). The initial and final areas of the solenoid's cross-section can be expressed as: \(A_i = \pi r_i^2\) and \(A_f =\pi r_f^2\) Since \(\frac{r_i}{r_f} = 14\), we can rewrite the final radius as \(r_f = \frac{r_i}{14}\). Now we can express the final area in terms of the initial area: \(A_f = \pi \left(\frac{r_i}{14}\right)^2 = \frac{1}{196}\pi r_i^2 = \frac{1}{196}A_i\)

05

Find the Final Magnetic Field

Now we can plug in the relationship between initial and final areas into the flux conservation equation: \(n A_i (1.0~\mathrm{T}) = n \left(\frac{1}{196}A_i\right) B_f\) We can divide both sides by \(n A_i\) and solve for \(B_f\): \(B_f = 196 (1.0~\mathrm{T})\) So, the final magnetic field produced is \(\boxed{196~\mathrm{T}}\).

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