Question

Equal current $I$ flowing in two wires $\text{PQ}$ and $\text{QR}$ according to figure, one end of both wire is at infinity. $\angle PQR$ equals $\theta .$ The magnetic field at point $\text{O}$ is


Here $\text{O}$ is at $r$ distance from $\text{Q.}$ Where $\text{QO}$ is bisector of angle $\angle PQR$

• A. $\frac{{\mu }_{0}I}{4\pi r}\sin \left(\frac{\theta }{2}\right)$
• B. $\frac{{\mu }_{0}I}{4\pi r}\cot \left(\frac{\theta }{2}\right)$
• C. $\frac{{\mu }_{0}I}{4\pi r}\tan \left(\frac{\theta }{2}\right)$
• D. $\frac{{\mu }_{0}i}{2\pi r}\left[\frac{1+\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\right]$

Answer 1

The correct option is D $\frac{{\mu }_{0}i}{2\pi r}\left[\frac{1+\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\right]$


The net field at point $\text{O}$ is,

$B_O=B_{PQ}+B_{RS}=B$

From symmetry we can say, $B_{PQ}=B_{RS}$

$\therefore B_O=2B....\left(1\right)$

Field due to a semi-infinite wire is,

$B=\frac{{\mu }_{0}I}{4\pi x}(\sin {\theta }_1+\sin {\theta }_2)[\because {\theta }_1={90}^{\circ }-\frac{\theta }{2}\text{and}{\theta }_2={90}^{\circ }]$

From the figure,

$\sin \frac{\theta }{2}=\frac{x}{r}\Rightarrow x=r\sin \frac{\theta }{2}$

From (1) we get,

$B_O=2\times \left(\frac{{\mu }_{0}I}{4\pi x}\right)[\sin (90-\frac{\theta }{2})+\sin {90}^{\circ }]$

$=2\times \frac{{\mu }_{0}i}{4\pi r\sin \frac{\theta }{2}}[1+\cos \frac{\theta }{2}]$

$B_O=\frac{{\mu }_{0}i}{2\pi r}\left[\frac{1+\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\right]$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.