Answer
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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.
Complete step by step answer:
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): $20A$
Additional Information
The magnetic field and electric field are similar in many ways. Both the magnetic field and electric field have a long-range. They both are inversely proportional to the square of the distance of the point from the source. Both fields apply the principle of superposition.
Note
The magnetic field is inversely proportional to the distance between the conductor and the point that we consider. This means that as the distance increases the magnetic field decreases. Hence we get that the magnetic field will be the maximum on the surface of the conductor. The value of permeability of free space should be noted.
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.
Complete step by step answer:
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): $20A$
Additional Information
The magnetic field and electric field are similar in many ways. Both the magnetic field and electric field have a long-range. They both are inversely proportional to the square of the distance of the point from the source. Both fields apply the principle of superposition.
Note
The magnetic field is inversely proportional to the distance between the conductor and the point that we consider. This means that as the distance increases the magnetic field decreases. Hence we get that the magnetic field will be the maximum on the surface of the conductor. The value of permeability of free space should be noted.
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