Physics - Magnetism Question with Solution | TestHub

(i) The magnetic field at the centre due to the curved portion $\mathrm{DA}=\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }R} \left(\frac{3\mathrm{\pi }}{2}\right)$

According to the right-hand screw rule, the magnetic field will be into the plane of the paper.



(ii) The magnetic field at $\mathrm{M}$ due to $\mathrm{AB}$ is zero.

(iii) The magnetic field at the centre due to the curved portion $BC$ is $\frac{{\mathrm{\mu }}_{0}i}{4\mathrm{\pi }2R} \left(\frac{\mathrm{\pi }}{2}\right).$

According to the right-hand screw rule, the magnetic field will be into the plane of the paper.

(iv) The magnetic field at $\mathrm{M}$ due to $\mathrm{DC}$ is zero.

Hence, the resultant magnetic field at $\mathrm{M}$

$B_M=B_{small loop}+B_{big loop}$

Both are perpendicular downward

$B_M=\frac{{\mu }_{O}i}{4\pi R}\left(\frac{3}{2}\pi \right)+\frac{{\mu }_{O}i}{4\pi \left(2R\right)}\left(\frac{\pi }{2}\right)$

$B_M=\frac{7}{16}\frac{{\mu }_{O}i}{R} ⨂$