Question

Figure shows an infinitely long current carrying wire bent at an angle $\alpha$. The magnetic induction at the point $\text{P}$ located on the angle bisector at a distance $x$ from the point $\text{O}$ at the bend is given by $\frac{{\mu }_{0}I}{2\pi x}\cot \frac{\alpha }{n}$. The value of $n$ is

Answer 1

The angle subtended by point $\text{P}$ at point $\text{O}$ is $\alpha /2$ and the other point is at infinity, so ${\theta }_2=0^{\circ }\left(\text{long wire}\right)$.

$\therefore$ We can use the expression for magnetic induction due to a finite wire by considering these angles.

The magnitude of magnetic induction at point $\text{P}$ due to the two wires is given as

$B_P=2\times B_{\text{due to one wire at P}}$
[$\because$ Magnetic field due to both wires are in same direction]

$B_P=2\times \frac{{\mu }_{o}I}{4\pi r}(\cos \alpha /2+\cos 0^{\circ })$

$B_P=2\times \frac{{\mu }_{o}I}{4\pi \left(x\sin \alpha /2\right)}(\cos \alpha /2+\cos 0^{\circ })$

$B_P=\frac{{\mu }_{o}I}{2\pi x\sin \alpha /2}(1+\cos \alpha /2)$

$B_P=\frac{{\mu }_{o}I}{2\pi x\sin \alpha /2}\times 2{\cos }^2\alpha /4$

$B_P=\frac{{\mu }_{o}I}{\pi x2\sin \alpha /4\cos \alpha /4}\times {\cos }^2\alpha /4$

$B_P=\frac{{\mu }_{o}I}{2\pi x}\cot \frac{\alpha }{4}$

$\therefore n=4$.

Note :- Be careful while putting the angles in the formula of the magnetic field due to a finite wire.