MHTCET 13 Apr 2026 Shift 2 Paper

Concept:Parallel current-carrying wires exert a magnetic force on each other, calculated using Ampere's force law for long straight conductors.Explanation:For two long parallel wires separated by distance $d$, the magnetic force on a length $l$ of either wire is given by:$$F = \frac{\mu_0 I_1 I_2 l}{2\pi d}$$Here, $I_1 = 2\ \mathrm{A}$, $I_2 = 4\ \mathrm{A}$, and $d = 20\ \mathrm{cm} = 0.20\ \mathrm{m}$.The length of wire considered is $l = 10\ \mathrm{cm} = 0.10\ \mathrm{m}$, and $\mu_0 = 4\pi \times 10^{-7}$ SI units.Substituting the given values into the formula:$$F = \frac{(4\pi \times 10^{-7})(2)(4)(0.10)}{2\pi(0.20)}$$Simplifying step by step:$$F = \frac{(4\pi \times 10^{-7})(8)(0.10)}{0.40\pi}$$$$F = \frac{3.2\pi \times 10^{-7}}{0.40\pi}$$$$F = 8 \times 10^{-7}\ \mathrm{N}$$Since both wires carry current in the same direction, the magnetic force between them is attractive.Thus, the magnitude of the force on either wire is $8 \times 10^{-7}\ \mathrm{N}$.Answer:The magnetic force on a length of $10\ \mathrm{cm}$ of either wire is $8 \times 10^{-7}\ \mathrm{N}$.Hence, the correct option is D: $8 \times 10^{-7}\ \mathrm{N}$.