MHTCET 16 May 2024 Shift 1 Paper

Concept:Magnetic fields produced by coils at their common centre are perpendicular vectors because the coils lie in perpendicular planes.So, the net field magnitude is found using the Pythagorean theorem.Explanation:For coil P, with radius $R$ and current $I$, the magnetic field at the centre is:$$B_P = \frac{\mu_0 I}{2R}$$For coil Q, with the same radius $R$ but current $\sqrt{8}I$, the field is:$$B_Q = \frac{\mu_0 \sqrt{8} I}{2R}$$Since $B_P$ and $B_Q$ are perpendicular, the net field is:$$B_{\text{net}} = \sqrt{B_P^2 + B_Q^2}$$Substituting the values:$$B_{\text{net}} = \sqrt{\left(\frac{\mu_0 I}{2R}\right)^2 + \left(\frac{\mu_0 \sqrt{8} I}{2R}\right)^2}$$Taking the common factor out:$$B_{\text{net}} = \frac{\mu_0 I}{2R} \sqrt{1 + 8}$$$$\Rightarrow B_{\text{net}} = \frac{\mu_0 I}{2R} \times 3 = \frac{3\mu_0 I}{2R}$$Answer:The magnitude of the magnetic field at the common centre is $\frac{3 \mu_0 I}{2R}$.Hence, the correct option is B.