Question

Five parallel infinite wires are placed at the vertices of a regular polygon. Four wires carry current $I_0$ each. While the fifth wire carries current $3I_0$ as shown. The resultant magnetic field at the centre O is :-

• A. 0
• B. $\frac{{\mu }_0{I}_0}{2\pi a}$
• C. $\frac{{\mu }_0{I}_0}{\pi a}$
• D. $\frac{{\mu }_{0}7I_0}{2\pi a}$

Answer 1

Answer: (C)

$\frac{{\mu }_0{I}_0}{\pi a}$

$Given:$ Four wires = $I_o$ each

Fifth wire= 3$I_o$

$Solution:$ Here $B_1$ = $B_2$ = $B_3$ = $B_4$ = $\frac{\mu oIo}{2\pi a}$ and $B_5$ = $\frac{\mu o3Io}{2\pi a}$ = $B_o$

Dissoaiate all the magnetic field along X - axis and Y - axix, we get

$B_x$ = $B_1$+$B_2$cos(72${}^o$) - $B_3$cos(36${}^o$) -$B_4$cos(36${}^o$)+$B_5$cos(72${}^o$)

= $B_o${ 1+4cos(72${}^o$) -2cos(36${}^o$) } = 0.62$B_o$

And

$B_y$ = $B_2$ sin(72${}^o$) + $B_3$sin(36${}^o$)-$B_4$sin(36${}^o$)-$B_5$sin(72${}^o$) = $B_o$(-2sin(72${}^o$) )

= -2$B_o$sin(72${}^o$) = -1.9$B_o$

$\therefore$ $Bnet$ = $\sqrt{B_{x}2+B_{y}2}$ = $B_o$$\sqrt{.38+3.62}$ =2$B_o$= $\frac{2\mu oIo}{2\pi a}$

$Bnet$ = $\frac{\mu oIo}{\pi a}$

$So,the$ $correct$ $option$ : $C$