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Consider an aluminum wire with 1 mm diameter below its critical temperature of 1.2 Kelvin. It's...

Question:

Consider an aluminum wire with 1 mm diameter below its critical temperature of 1.2 Kelvin. It's critical magnetic field is 0.01 Tesla.

a. What is the maximum current in amps that you can send down this wire for it to remain superconducting?

Current:

The electrical quantity illustrates that the electric charge moves in the conducting material in unit time known as current. The ampere is a measurable unit to define the current.

Answer and Explanation:


Given Data:

  • The diameter of aluminum wire is: {eq}{d_a} = 1\;{\rm{mm}} = 0.001\;{\rm{m}} {/eq}
  • The critical magnetic field is: {eq}{B_a} = 0.01\;{\rm{Tesla}} {/eq}


The maximum current send down aluminum wire for it to remain superconducting is: {eq}{I_a} {/eq}

The permeability of free space is: {eq}{\mu _o} = 4\pi \times {10^{ - 7}}\;{\rm{N/}}{{\rm{A}}^2} {/eq}


The expression for critical magnetic field for aluminum wire carry current is

{eq}{B_a} = \dfrac{{{\mu _o}{I_a}}}{{2\pi \left( {\dfrac{{{d_a}}}{2}} \right)}} {/eq}


Substitute the value and solve the above expression

{eq}\begin{align*} 0.01\;{\rm{T}} &= \dfrac{{4\pi \times {{10}^{ - 7}}\;{\rm{N/}}{{\rm{A}}^2} \times {I_a}}}{{2\pi \left( {\dfrac{{0.001\;{\rm{m}}}}{2}} \right)}}\\ {I_a} &= \dfrac{{0.01\;{\rm{T}} \times 0.001\;{\rm{m}}}}{{4 \times {{10}^{ - 7}}\;{\rm{N/}}{{\rm{A}}^2}}}\\ &= 25\;{\rm{A}} \end{align*} {/eq}


Thus the maximum current send down aluminum wire for it to remain superconducting is {eq}25\;{\rm{A}} {/eq}.


Learn more about this topic:

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What is Electric Current? - Definition, Unit & Types

from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 6 / Lesson 7
310K

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