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# A copper wire of diameter $1.6mm$ carries a current $i$. The maximum magnetic field due to this wire is $5 \times {10^{ - 3}}T$. The value of $i$ is(A) $40A$(B) $5A$(C) $20A$(D) $2A$

Last updated date: 20th Jun 2024
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Hint In the question, we have a copper wire through which a current is flowing. The diameter of the wire is given so that we can find the radius of the wire. There will be a magnetic field generated due to the moving current in the wire. The maximum magnetic field is also given. We have to find the current through the conductor.

Formula used
$B = \dfrac{{{\mu _0}i}}{{2\pi r}}$
Where $B$ stands for the magnetic field due to the current through the wire, $i$ stands for the current flowing through the conductor, $r$ stands for the radius of the wire, ${\mu _0}$ is a constant called the permeability of free space.

According to Biot-Savart’s law,
The magnetic field due to an infinitely long current carrying conductor can be written as,
$B = \dfrac{{{\mu _0}i}}{{2\pi r}}$
From this, we get
$i = \dfrac{{2\pi rB}}{{{\mu _0}}}$
The maximum magnetic field is given as,
$B = 5 \times {10^{ - 3}}T$
The diameter of the wire is given by,
$d = 1.6mm$
From this, we can find the radius as,
$r = \dfrac{d}{2} = \dfrac{{1.6}}{2} = 0.8mm$
This will become,
$r = 0.8 \times {10^{ - 3}}m$
The permeability of free space can be written as,${\mu _0} = 4\pi \times {10^{ - 7}}$
Substituting, $B = 5 \times {10^{ - 3}}T$, $r = 0.8 \times {10^{ - 3}}m$ and ${\mu _0} = 4\pi \times {10^{ - 7}}$ in the equation $i = \dfrac{{2\pi rB}}{{{\mu _0}}}$
We get,
$i = \dfrac{{2\pi \times \left( {0.8 \times {{10}^{ - 3}}} \right) \times \left( {5 \times {{10}^{ - 3}}} \right)}}{{4\pi \times {{10}^{ - 7}}}} = 20A$
$\therefore i = 20A$

The answer is: Option (C): $20A$