Physics - Magnetism Question with Solution | TestHub

(A) Magnetic field at the plane of the ring is perpendicular to the plane. However, they bend as they move forward.(B) By symmetry, we can say that$\overrightarrow{B}$will be same at all the points having the same radial distance. so$\overrightarrow{B}$$\left(x,y\right)$will depend only the radial distance$r=\sqrt{x^2+y^2}$(C)${\left(B_{\mathrm{net} }\right)}_{\mathrm{centre} }=\frac{{\mu }_0{I}_2}{2\left(2\mathrm{R}\right)}-\frac{{\mu }_0{I}_1}{2\mathrm{R}}=\frac{{\mu }_0}{4\mathrm{R}}\left(I_2-2I_1\right)$Since,$I_2>2I_1$so$B_{\mathrm{net}}$at the centre will be non-zero in$\otimes$direction, But at some other point,$B_{\mathrm{net}}$may be zero. From the graph, it is clear that$B_{\mathrm{net}}=0$for some value of$r\in \left(0,R\right)$. Sooption (C) is incorrect(D) For the graph, it is clear that$B=-ve\Rightarrow$in$\otimes$direction for$r\in \left(R \mathrm{to} 2\mathrm{R}\right)$, so option D is also incorrect.