Question

In the given figure, what is the magnetic field induction at point O.

• A. $\frac{{\mu }_{0}l}{4\pi r}$
• B. $\frac{{\mu }_{0}l}{4r}+\frac{{\mu }_{0}l}{2\pi r}$
• C. $\frac{{\mu }_{o}l}{4r}+\frac{{\mu }_{0}l}{4\pi r}$
• D. $\frac{{\mu }_{0}l}{4r}-\frac{{\mu }_{0}l}{4\pi r}$

Answer 1

The correct option is C $\frac{{\mu }_{o}l}{4r}+\frac{{\mu }_{0}l}{4\pi r}$

Magnetic field due to straight wire above O is zero
i.e., $B_1=0$(since, $\theta =0^o$)
The magnetic field due to semi circular part
$B_2=\left(\frac{{\mu }_{0}nI}{2r}\right)=\frac{{\mu }_0\left(1/2\right)I}{2r}=\frac{{\mu }_{0}I}{4r}$
The magnetic field due to lower straight portion
$B_3=\frac{1}{2}\frac{{\mu }_{0}I}{2\pi r}$(upward)
Net magnetic field $B=B_1+B_2+B_3$
$=0+\frac{{\mu }_{0}I}{4r}+\frac{{\mu }_{0}I}{4\pi r}$(upwards).