Question

Two semi-infinitely long straight current carrying conductors are in form of an $L-$shape as shown in figure. The common end is at the origin. What is the value of magnetic field at a point $(a,b)$, if both the conductors carrying same current $I$?

• A. $\frac{{\mu }_{0}I}{4\pi ab}[1+\frac{a}{\sqrt{a^2+b^2}}]$
• B. $\frac{{\mu }_{0}I}{4\pi ab}[(a+b)+\sqrt{a^2+b^2}]$
• C. $\frac{{\mu }_{0}I}{4\pi a}[\frac{b}{\sqrt{a^2+b^2}}]$
• D. $\frac{{\mu }_{0}I}{4\pi ab}[1+\frac{\sqrt{a^2+b^2}}{a+b}]$

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{4\pi ab}[(a+b)+\sqrt{a^2+b^2}]$

For the wire along $x-axis$, the magnetic field produced is



$$B_1=\frac{{\mu }_{0}I}{4\pi d}(\sin {\varphi }_1+\sin {\varphi }_2)...\left(1\right)$$Here, ${\varphi }_1={90}^{\circ }$, ${\varphi }_2={90}^{\circ }-{\theta }_2$, and $d=b$

$\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi d}[\sin {90}^{\circ }+\sin ({90}^{\circ }-{\theta }_2)]$

$\Rightarrow B_1=\frac{{\mu }_{0}I}{4\pi d}[1+\cos {\theta }_2]$

From the geometry:
$\cos {\theta }_2=\frac{a}{\sqrt{a^2+b^2}}$

$\therefore B_1=\frac{{\mu }_{0}I}{4\pi d}[1+\frac{a}{\sqrt{a^2+b^2}}]$

Now, for the wire $AB$ along, $y-axis$, the magnetic filed produced at $P$ is,

$B_2=\frac{{\mu }_{0}I}{4\pi d}[\sin {\varphi }_1+\sin {\varphi }_2]$

Here, $d=a$, ${\varphi }_1={90}^{\circ }-{\theta }_1$ and ${\varphi }_2={90}^{\circ }$

$\Rightarrow B_2=\frac{{\mu }_{0}I}{4\pi d}[\sin ({90}^{\circ }-{\theta }_1)+\sin {90}^{\circ }]$

$\Rightarrow B_2=\frac{{\mu }_{0}I}{4\pi d}[\cos {\theta }_1+1]$

From the geometry, substituting for $\cos {\theta }_1$,
$\cos {\theta }_1=\frac{b}{\sqrt{a^2+b^2}}$

$\Rightarrow B_2=\frac{{\mu }_{0}I}{4\pi d}[1+\frac{b}{\sqrt{a^2+b^2}}]$

The direction of field $B_1$ and $B_2$ are perpendicularly inwards to plane i.e., $-ve$ $z-axis$.

Hence, net magnetic field,
$B_P=B_1+B_2$

$\Rightarrow B_P=\frac{{\mu }_{0}I}{4\pi d}[1+\frac{a}{\sqrt{a^2+b^2}}]+\frac{{\mu }_{0}I}{4\pi d}[1+\frac{b}{\sqrt{a^2+b^2}}]$

$\Rightarrow B_P=\frac{{\mu }_{0}I}{4\pi }[\frac{a^2+b^2}{ab\left(\sqrt{a^2+b^2}\right)}]+\frac{{\mu }_{0}I}{4\pi }[\frac{1}{b}+\frac{1}{a}]$

$\Rightarrow B_P=\frac{{\mu }_{0}Iab}{4\pi }[\sqrt{a^2+b^2}+(a+b)]$

Hence, option (b) is correct answer.