Question

In the figure, what is the magnetic field at the point $O$?

• A. $\frac{{\mu }_{o}I}{4r}-\frac{{\mu }_{o}I}{4\pi r}$
• B. $\frac{{\mu }_{o}I}{4r}+\frac{{\mu }_{o}I}{4\pi r}$
• C. $\frac{{\mu }_{o}I}{4\pi r}+\frac{{\mu }_{o}I}{2\pi r}$
• D. $\frac{{\mu }_{o}I}{4\pi r}$

Answer 1

The correct option is B $\frac{{\mu }_{o}I}{4r}+\frac{{\mu }_{o}I}{4\pi r}$

Field due to straight wire $1$ at $O$ will be zero because $O$ lies on the axis of wire $1$

So,
$B_1=0$
Field due to semi-circular wire at $O$

$B_2=\frac{{\mu }_{0}i\theta }{4\pi r}=\frac{{\mu }_{0}i\left(\pi \right)}{4\pi r}⨀$

$\Rightarrow B_2=\frac{{\mu }_{0}i\left(\pi \right)}{4\pi r}=\frac{{\mu }_{0}i}{4r}⨀$

Direction of field can be checked by Ampere's right hand thumb rule.

Field at $O$ due to semi-infinite wire $3$

$B_3=\frac{{\mu }_{0}i\theta }{4\pi r}=(\sin {\varphi }_1+\sin {\varphi }_2)$

$B_3=\frac{{\mu }_{0}i\theta }{4\pi r}=(\sin {90}^0+\sin 0^0)$

$B_3=\frac{{\mu }_{0}i}{4\pi r}⨀$

Resultant field at $O$,

$B_{net}=B_1+B_2+B_3$​​​​

($B_3$​​​ and $B_2$ are in same direction)​​​

$\Rightarrow B_{net}=0+\frac{{\mu }_{0}i}{4r}+\frac{{\mu }_{0}i}{4\pi r}$

$\Rightarrow B_{net}=\frac{{\mu }_{0}i}{4r}+\frac{{\mu }_{0}i}{4\pi r}$