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A long solenoid that has 1.00 times 10^3 turns uniformly distributed over a length of 0.400 m...

Question:

A long solenoid that has {eq}1.00 \times 10^3 {/eq} turns uniformly distributed over a length of {eq}0.400 \ m {/eq} produces a magnetic field of magnitude {eq}1.00 \times 10^{-4} \ T {/eq} at it's centre. What current is required in the windings for that to occur?

Electric Current:

In physics, the term which is utilized to represent the flow rate of an electron from one point to another point is known as electric current. In the International Standard of the unit system, the electric magnitude of the current is described by Ampere (A).

Answer and Explanation:


Given data

  • The number of turns of the long solenoid is {eq}n = 1 \times {10^3}\;{\rm{turns}} {/eq}
  • The length of the solenoid is {eq}l = 0.4 m {/eq}
  • The magnitude of the magnetic field is {eq}B = 1 \times {10^{ - 4}}\;{\rm{T}} {/eq}


Note- The permeability of the free space is {eq}{\mu _0} = 4\pi \times {10^{ - 7}}\;{\rm{T}} \cdot {\rm{m/A}} {/eq}

The required current is calculated as,

{eq}\begin{align*} I &= \dfrac{{Bl}}{{{\mu _0}n}}\\ &= \dfrac{{\left( {1 \times {{10}^{ - 4}}\;{\rm{T}}} \right)\left( {0.4\;{\rm{m}}} \right)}}{{\left( {4\pi \times {{10}^{ - 7}}\;{\rm{T}} \cdot {\rm{m/A}}} \right)\left( {1 \times {{10}^3}} \right)}}\\ &= 0.032\;{\rm{A}} \end{align*} {/eq}


Thus, the electric current is 0.032 A.


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