Physics - Magnetism Question with Solution | TestHub

The magnetic field due to a finite straight wire carrying current $I$ at a perpendicular distance $r$ is given by $B=\frac{{\mu }_{0}I}{4\pi r}(\sin {\theta }_1+\sin {\theta }_2)$.

Since the point P lies on the perpendicular bisector of the wire of length $2L$, the angles subtended by the ends of the wire at P are equal, so ${\theta }_1={\theta }_2=\theta$.

From the geometry of the figure, $\sin \theta =\frac{L}{\sqrt{r^2+L^2}}$.

Substituting these values into the formula, we get:

$B=\frac{{\mu }_{0}I}{4\pi r}(2\sin \theta )=\frac{{\mu }_{0}I}{2\pi r}\times \frac{L}{\sqrt{r^2+L^2}}$

Answer: $\frac{{\mu }_{0}I}{2\pi r}\times \frac{L}{\sqrt{r^2+L^2}}$