Question

Infinite number of straight wires each carrying current i are equally spaced that justin wires have current in opposite directions that magnetic field at o is

Answer 1

க்ஷுக்ஷ்ப்ய்க்ஷ்ய் ய்8ப்குப்க்ஷுக் ப்ப்ஹ்ய்ட்

Answer 2

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Magnetic field from a straight finite length wire is  $\mu I\frac { \left( \sin { { \theta }_{ 1 }+\sin { { \theta }_{ 2 } } } \right) }{ 4\pi d }$

d= perpendicular distance of point from wire

I = Current, ${ \theta }_{ 1 }\quad and\quad { \theta }_{ 2 }$are angle of elevation from the point to top and bottom of wire

$d\quad =\quad a\cos { 30 } =a\frac { \sqrt { 3 } }{ 2 }$

So the magnetic field at point P from the wires are given as $\frac { \frac { \mu I }{ 4\pi a{ \left( 3 \right) }^{ 1/2 } } }{ 2 } -\quad \frac { \frac { \mu I }{ 4\pi 2a{ \left( 3 \right) }^{ 1/2 } } }{ 2 } +\frac { \frac { \mu I }{ 4\pi 3a{ \left( 3 \right) }^{ 1/2 } } }{ 2 }$

$\mu I2\quad \frac { \left( \frac { 1-1 }{ 2 } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } ........ \right)  }{ 4\pi a{ \left( 3 \right) }^{ 1/2 } }$

$\log { \left( 1+x \right) } =\quad \frac { x-{ x }^{ 2 } }{ 2 } +\frac { { x }^{ 3 } }{ 3 } -\frac { { x }^{ 4 } }{ 4 } ........$

If we consider x=1 then we get $\log { 2\quad =\quad 1-\frac { 1 }{ 2 } } +\frac { 1 }{ 3 } -\frac { 1 }{ 4 } ......$

Therefore the magnetic field at point P is equal to $\quad \frac { \mu I2\log { 2 } }{ 4\pi a{ \left( 3 \right) }^{ 1/2 } }$