Question

Magnetic field at point $P$ due to following current distribution is :-

• A. $\frac{{\mu }_{0}I}{{\pi r}^2}\sqrt{r^2-a^2}$
• B. $\frac{{\mu }_{0}Ia}{{\pi r}^2}$
• C. $\frac{{\mu }_{0}I}{{2\pi r}^2}\sqrt{r^2-a^2}$
• D. $\frac{{\mu }_{0}Ia}{{2\pi r}^2}$

Answer 1

The correct option is B $\frac{{\mu }_{0}Ia}{{\pi r}^2}$

$|{\overrightarrow{B}}_1|=|{\overrightarrow{B}}_2|=\frac{{\mu }_{0}I}{4\pi R}\times 2$

$=\frac{{\mu }_{0}I}{2\pi R}=\frac{{\mu }_{0}I}{2\pi r}$

Angle between ${\overrightarrow{B}}_1$ and ${\overrightarrow{B}}_2=180-2\theta$

$\Rightarrow B_{net}=\sqrt{B_1^2+B_2^2+2B_1{B}_2\cos (180-2\theta )}$

$=\frac{{\mu }_{0}I}{2\pi r}\sqrt{2}\times \sqrt{1-\cos \left(2\theta \right)}=\frac{{\mu }_{0}I}{\sqrt{2}\pi r}\sqrt{2}\sqrt{{\sin }^2\left(\theta \right)}$

$\Rightarrow B_{net}=\frac{{\mu }_{0}I}{\pi r}\times \frac{a}{r}=\frac{{\mu }_{0}Ia}{\pi r^2}$

Therefore option B is correct.