Question

Four very long current carrying wires in the same intersect to form a square $40.0cm$ on each side as shown in the figure. What is the magnitude of current $I$ so that the magnetic field at the centre of the square is zero?

• A. $22A$
• B. $38A$
• C. $2A$
• D. $18A$

Answer 1

The correct option is C $2A$

Magnitude of the magnetic field
At a perpendicular distance $x$ from infinite long wire is given by
$B=\frac{\mu d}{2\pi x}$
For wire $AB{B}_1\frac{20{\mu }_0}{2\pi x}\hat{k}....\left(i\right)$
Due to $BC$
$B_2=\frac{{\mu }_{0}l}{2\pi x}(-\hat{k})....\left(ii\right)$
Due to $DC$
$B_3=\frac{8{\mu }_0}{2\pi x}(-\hat{k})...,.\left(iii\right)$
Due to $AD$
$B_4=\frac{10{\mu }_0}{2\pi x}(-\hat{k})....\left(iv\right)$
From Eqs. (i), (ii), (iii) and (iv)
$B_1+B_2+B_3+B_4=0$ [From question]
$\Rightarrow \frac{20{\mu }_0}{2\pi x}-\frac{{\mu }_{0}l}{2\pi x}=\frac{8{\mu }_0}{2\pi x}+\frac{10{\mu }_0}{2\pi x}$
$\Rightarrow 20-l=8+10$
$\Rightarrow l=20-18$
$l=+2A$
(Positive sign shows the direction of current is same direction as that of given direction)