Answered: or the future Electron Ion Collider…

Related questions

Question

or the future Electron Ion Collider (EIC) a proposed magnet for the detector is a solenoid with the following properties:

B = 3 Tesla (assume uniform inside, 0 outside)

Length L = 4.0 m

Inner Radius R = 1.5 m

Use the long thin solenoid approximation of uniform field, even though this is a short squat solenoid

A). Assuming the coil has radius R = 1.5 m, what is the current density (Amp per unit length)

B) Calculate the total stored energy in the magnetic field, in Joules.
Given that 1 gallon of gasoline contains about 10 ⁸ Joules of chemical energy is this a large, medium, or small amount of stored energy?

C) The coil would be superconducting. If the coil were to suddenly lose it's superconductivity and the magnetic field ramped down extremely rapidly, where would the stored energy go?

Expert Solution

Step 1

solenoid with magnetic field= 3T

Length= 4.0m

Inner radius=1.5m

$u \sin g a m p e r e^{'} s c i r c u i t a l l a w \oint \vec{B} . \vec{d l} = μ_{0} I_{t o t} (1) I_{t o t} = n L I (2) w h e r e n i s t h e n u m b e r o f t u r n s p e r u n i t l e n g t h L i s t h e l e n g t h o f t h e s o l e n o i d f r o m e q u a t i o n (1) a n d (2) B L = μ_{0} n L I n I = J = t h e r e q u i r e d c u r r e n t d e n s i t y p e r u n i t l e n g t h B L = μ_{0} L J J = \frac{B}{μ_{0}} = \frac{3}{4 π \times 10^{- 7}} A / l e n g t h = 0.24 \times 10^{7} A m p e r e / l e n g t h$

Answer: current density per unit length= $0.24 \times 10^{7} A m p e r e / m e t r e$

Step 2

Energy stored in the magnetic field

$E = \frac{B^{2}}{2 μ_{0}} \times A L A = {πr}^{2} = 3.14 \times {(1.5)}^{2} m^{2} = 7.065 m^{2} L = 4 m, B = 3 Tesla E = \frac{{(3)}^{2} \times 7.065 \times 4}{2 \times 2 π \times 10^{- 7}} = 2.025 \times 10^{8} joule$

Answer: energy stored in the magnetic field= $2.025 \times 10^{8} J o u l e$

Step by step

Solved in 3 steps