- #1
arpon
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Suppose there is a potential difference between points A and B which are connected by a straight wire. The current in AB is [itex]i[/itex].
We want to calculate the magnetic field at point C which is at distance [itex]r[/itex] from the middle point P of the wire and CP is perpendicular to AB.
At first, we use Ampere's law. We consider a circular loop (of radius [itex]r[/itex]) around the wire. Because of symmetry, magnetic field is always same in this loop. As the current through this loop is [itex]i[/itex],we get,
[itex]\oint \vec B \cdot d \vec s = B(2\pi r) = \mu _0 i[/itex]
So, [itex]B = \frac {\mu _0 i}{2\pi r}[/itex]
But, now we use Biot-Savart's law.
[itex]dB = \frac {\mu _0 i}{4\pi } \frac {i dl sin\theta }{r^2 + l^2}[/itex]
So, [itex]B = \frac {\mu _0 i}{4\pi r} \frac {L}{\sqrt {r^2 + (\frac{L}{2})^2}} [/itex], where [itex]L = [/itex]the length of AB.
We want to calculate the magnetic field at point C which is at distance [itex]r[/itex] from the middle point P of the wire and CP is perpendicular to AB.
At first, we use Ampere's law. We consider a circular loop (of radius [itex]r[/itex]) around the wire. Because of symmetry, magnetic field is always same in this loop. As the current through this loop is [itex]i[/itex],we get,
[itex]\oint \vec B \cdot d \vec s = B(2\pi r) = \mu _0 i[/itex]
So, [itex]B = \frac {\mu _0 i}{2\pi r}[/itex]
But, now we use Biot-Savart's law.
[itex]dB = \frac {\mu _0 i}{4\pi } \frac {i dl sin\theta }{r^2 + l^2}[/itex]
So, [itex]B = \frac {\mu _0 i}{4\pi r} \frac {L}{\sqrt {r^2 + (\frac{L}{2})^2}} [/itex], where [itex]L = [/itex]the length of AB.