Question

A current carrying wire AB with current $i_0$ is placed as shown in Figure. Find the magnetic field at the origin.

• A. $0.01\frac{{\mu }_0{i}_0}{\pi }$
• B. $0.15\frac{{\mu }_0{i}_0}{\pi }$
• C. $0.49\frac{{\mu }_0{i}_0}{\pi }$
• D. $\frac{{\mu }_0{i}_0}{\pi }$

Answer 1

The correct option is B $0.15\frac{{\mu }_0{i}_0}{\pi }$

$x$ is the perpendicular distance of wire AB from the origin

$\cos \theta =\frac{x}{OA}$

or, $x=\cos \theta \times 3$

Now, $\cos \theta =\frac{4}{5}\left(from△OAB\right)$

$\therefore x=\frac{4}{5}\times 3=\frac{12}{5}$

Hence magnetic field

$B=\frac{{\mu }_0{I}_0}{4\pi x}[\sin \theta +\sin (90-\theta \left)\right]$

$=\frac{{\mu }_0{I}_0}{4\pi \times \frac{12}{5}}[\frac{3}{5}+\frac{4}{5}]$

$=\frac{5}{4\pi \times 12}\times \frac{7}{5}{\mu }_0{I}_0=0.1458\frac{{\mu }_0{I}_0}{\pi }$

$=0.15\frac{{\mu }_0{I}_0}{\pi }$