Physics - Magnetism Question with Solution | TestHub

For a wire in the x-y plane:

$$\mathrm{d}\vec{\mathrm{B}}=\frac{{\mu }_0}{4\pi }\frac{\mathrm{id}\vec{\mathrm{l}}\times (-\vec{\mathrm{r}})}{|-\vec{\mathrm{r}}{|}^3}$$

$$\mathrm{d}\vec{\mathrm{B}}=-\frac{{\mu }_0}{4\pi }\frac{\mathrm{i}\overrightarrow{\mathrm{dl}}\times \vec{\mathrm{r}}}{|-\vec{\mathrm{r}}{|}^3}$$

$$|\mathrm{dB}|=\frac{{\mu }_0\mathrm{i}}{4\pi }\frac{|\mathrm{dl}|\times |\vec{\mathrm{r}}|\sin \alpha }{|\vec{\mathrm{r}}{|}^3}$$

$$|\mathrm{dB}|=\frac{{\mu }_0}{4\pi }\frac{\mathrm{ird}\theta }{{\mathrm{r}}^2}=\frac{{\mu }_0\mathrm{i}}{4\pi \mathrm{r}}\mathrm{d}\theta$$

(A) In the given situation, all current elements will produce $\vec{\mathrm{B}}$ in the same direction.

$$B_{net}=\frac{{\mu }_0}{4\pi }\int \frac{I\mathrm{d}\theta }{r}=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{\pi /2}\frac{\mathrm{d}\theta }{1+\cos \theta }$$

$$B_{net}=\frac{{\mu }_{0}I}{4\pi }{\int }_0^{\pi /2}\frac{1}{1+\cos \theta }\mathrm{d}\theta =\frac{{\mu }_{0}I}{4\pi }{\int }_0^{\pi /2}\frac{1}{2}{\sec }^2\left(\frac{\theta }{2}\right)\mathrm{d}\theta$$

$$B_{net}=\frac{{\mu }_{0}I}{4\pi }{\left(\frac{1}{2}\frac{\tan \left(\frac{\theta }{2}\right)}{\frac{1}{2}}\right)}_0^{\pi /2}=\frac{{\mu }_{0}I}{4\pi }$$

(B) $\mathrm{r}=4\sec \theta$

$$B_{net}={\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}\frac{{\mu }_{0}I}{4\pi }\frac{\mathrm{d}\theta }{4\sec \theta }=\frac{{\mu }_{0}I}{16\pi }{\int }_{-\frac{\pi }{2}}^{+\frac{\pi }{2}}\cos \theta \mathrm{d}\theta =\frac{{\mu }_{0}I}{16\pi }(2)$$

(C) $\vec{B}$ due to a moving charge is $\vec{B}=\frac{{\mu }_0}{4\pi }\frac{\mathrm{q}\vec{v}\times \vec{r}}{|\vec{r}{|}^3}$. If $|\vec{r}\times \mathrm{m}\vec{v}|$ is constant, then $r$ may vary, so $|\vec{B}|$ may not be constant.

(D) This is a straight wire at a distance $r\cos \theta =a$ from the origin.