Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are i₁ = 10 A, i₂ = -13 A, i₃ = 10 A, i₄ = 7 A, i₅ = -12 A and i₆ = 18 A. The magnetic induction at a perpendicular distance of 10 cm from the cable is (μ₀ = 4π × 10⁻⁷ SI units) (a) 40 μ T (b) 375 μ T (c) 30 μ T (d) 35 μ T

Answer: (a). Outside the cable it acts like one wire carrying the net current (Ampere's law). Inet = 10 - 13 + 10 + 7 - 12 + 18 = 20 A B = (μ₀ I)/(2π r) = 2 × 10⁻⁷ × 200.1 = 4 × 10⁻⁵ T So the answer is (a) 40 μ T.

Six very long insulated copper wires are bound together to form a… | MHT CET 2026 (13 May, Shift 1)

Q1MHT CET 2026 · 13 MAY · SHIFT 1Easy

Ampere's Law, Cylinders, Solenoid & Toroid

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are $i_1 = 10$ A, $i_2 = -13$ A, $i_3 = 10$ A, $i_4 = 7$ A, $i_5 = -12$ A and $i_6 = 18$ A. The magnetic induction at a perpendicular distance of $10$ cm from the cable is $(\mu_0 = 4\pi \times 10^{-7}$ SI units $)$

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Answer: (a)

Outside the cable it acts like one wire carrying the net current (Ampere's law).
$I_{net} = 10 - 13 + 10 + 7 - 12 + 18 = 20$ A
$B = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 20}{0.1} = 4 \times 10^{-5}$ T
So the answer is (a) $40\ \mu$ T.

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Ampere's Law, Cylinders, Solenoid & Toroid question from MHT CET 2026 (13 May, Shift 1)

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