Question

A current carrying wire in the form of 'V' alphabet is kept as shown in the figure. Magnetic field intensity at point $P$ which lies on the angular bisector of V is

• A. $\frac{{\mu }_{0}i}{4\pi r_0}[1-\cos \alpha ]$
• B. $\frac{{\mu }_{0}i}{2\pi r_0}[1-\cos \alpha ]$
• C. $\frac{{\mu }_{0}i}{4\pi r_0}\frac{[1-\cos \alpha ]}{\sin \alpha }$
• D. $\frac{{\mu }_{0}i}{2\pi r_0}\frac{[1-\cos \alpha ]}{\sin \alpha }$

Answer 1

The correct option is D $\frac{{\mu }_{0}i}{2\pi r_0}\frac{[1-\cos \alpha ]}{\sin \alpha }$

Following from Bio-Savart Law, the magnetic field due to a finite current carrying wire at a point distance $d$ away from it is given by

$B=\frac{{\mu }_{0}i}{4\pi d}(sin{\theta }_1-sin{\theta }_2)$

where ${\theta }_1,{\theta }_2$ are angles that the line joining end point of wire to the point form with the line of projection from the point to the wire.

Here magnetic field due to each wire $=\frac{{\mu }_{0}i}{4\pi r_0}(sin{90}^{\circ }-sin({90}^{\circ }-\alpha \left)\right)$

$=\frac{{\mu }_{0}i}{4\pi r_0}(1-cos\alpha )$ in the out of the plane direction.

Hence net magnetic field at the point $P=\frac{{\mu }_{0}i}{2\pi r_0}(1-cos\alpha )$