Physics - Magnetism Question with Solution | TestHub

Current density in the remaining part: $J=\frac{I}{\pi R^2-\pi (R/3{)}^2}=\frac{9I}{8\pi R^2}$

Use superposition: field = field of the complete cylinder (density J) minus field of the removed cylinder (density J).

From the top view, the removed cylinder is internally tangent to the big one, so the separation of the axes is $d=R-\frac{R}{3}=\frac{2R}{3}$

A standard result: the field inside the empty cavity is uniform, $B=\frac{{\mu }_{0}Jd}{2}=\frac{{\mu }_0}{2}\cdot \frac{9I}{8\pi R^2}\cdot \frac{2R}{3}=\frac{3{\mu }_{0}I}{8\pi R}$

Point A lies inside the cavity (OA = R/2), so $B_A=\frac{3{\mu }_{0}I}{8\pi R}$ → (C) is correct.

Point B is the axis of the cavity: $B_B=\frac{{\mu }_{0}J(2R/3)}{2}-0=\frac{3{\mu }_{0}I}{8\pi R}$ → (A) is correct.

Hence (A) and (C).