Question

If the magnetic field at ‘P’ can be written as $K$ $\tan \left(\frac{\alpha }{2}\right)$, then $K$ is

• A. $\frac{{\mu }_{0}I}{4\pi d}$
• B. $\frac{{\mu }_{0}I}{2\pi d}$
• C. $\frac{{\mu }_{0}I}{\pi d}$
• D. $\frac{2{\mu }_{0}I}{4\pi d}$

Answer 1

The correct option is B $\frac{{\mu }_{0}I}{2\pi d}$

Magnetic field at point P will be due to two elements AC and AB which will be in same direction as out of the plane as shown in diagram.

We know that magnetic field due to current carrying wire is
$B=\frac{{\mu }_{0}I}{4\pi d}(\sin {\theta }_1+\sin {\theta }_2)$

So for wire AC, ${\theta }_1=-(90-\alpha )$
${\theta }_2=90$
$\therefore B_{AC}=\frac{{\mu }_{0}I}{4\pi d\sin \alpha }\left(\sin \right(-(90-\alpha )+\sin 90)$

$\therefore B_{AC}=\frac{{\mu }_{0}I}{4\pi d\sin \alpha }(1-\cos \alpha )$
$\therefore B_{AC}=\frac{{\mu }_{0}I}{4\pi d}\left(\tan \frac{\alpha }{2}\right)$

Now net magnetic field at point P will be double of the magnetic field due to element AC due to symmetry.
$\therefore B_P=\frac{{\mu }_{0}I}{2\pi d}\left(\tan \frac{\alpha }{2}\right)$

$\therefore K=\frac{{\mu }_{0}I}{2\pi d}$