Physics - Magnetism Question with Solution | TestHub

The magnetic field at the focus P of the parabola is given by the Biot-Savart law:

$$\overrightarrow{\mathrm{B}}=\frac{{\mu }_{\mathrm{o}}}{4\pi }\int \frac{\mathrm{Id}\vec{\ell }\times \overrightarrow{\mathrm{r}}}{{\mathrm{r}}^3}$$

Where $\overrightarrow{\mathrm{r}}$ and $\mathrm{Id}\vec{\ell }$ are as shown in figure. The direction of $\overrightarrow{\mathrm{B}}$ is perpendicular to the plane of the parabola and out of it. We note that $\frac{|\mathrm{d}\vec{\ell }\times \overrightarrow{\mathrm{r}}|}{{\mathrm{r}}^3}=\frac{\mathrm{rd}\ell \sin \alpha }{{\mathrm{r}}^3}=\frac{\mathrm{rd}\theta }{{\mathrm{r}}^2}=\frac{\mathrm{d}\theta }{\mathrm{r}}$ Therefore, $\mathrm{B}=\left({\mu }_{\mathrm{o}}\frac{\mathrm{I}}{4\pi }\right){\int }_0^{2\pi }\left(\frac{\mathrm{d}\theta }{\mathrm{r}}\right)$ The equation of the parabola is $\mathrm{r}(1-\cos \theta )=2\mathrm{a}$ So that, $B=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{2\pi }\frac{(1-\cos \theta )}{2a}d\theta =\frac{{\mu }_{0}I}{4a}=\frac{4\times \pi \times 10^{-7}\times 20}{4\times 10^{-3}}=2\pi \times 10^{-3}$ tesla