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 Wb/A-m) (a) 40 (b) 37.5 (c) 30 (d) 35

Answer: (a). Outside the cable, only the net current matters (Ampere's law). Inet = 10 - 13 + 10 + 7 - 12 + 18 = 20 A B = (μ₀ I)/(2π r) = 2 × 10⁻⁷ × 200.1 = 40 × 10⁻⁶ T So the answer is (a) 40 μ T.

Six very long insulated copper wires are bound together to form a… | MHT CET 2019 (3 May, Shift 1)

Q50MHT CET 2019 · 3 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 $I_1 = +10\text{ A}, I_2 = -13\text{ A}, I_3 = +10\text{ A}, I_4 = +7\text{ A}, I_5 = -12\text{ A}$ and $I_6 = +18\text{ A}$ . The magnetic induction at a perpendicular distance of 10 cm from the cable is

$\left(\mu_0 = 4\pi \times 10^{-7}\text{ Wb/A-m}\right)$

Show answer and solution

Answer: (a)

Outside the cable, only the net current matters (Ampere's law).
$I_{net} = 10 - 13 + 10 + 7 - 12 + 18 = 20$ A
$B = \frac{\mu_0 I}{2\pi r} = \frac{2 \times 10^{-7} \times 20}{0.1} = 40 \times 10^{-6}$ T
So the answer is (a) $40\ \mu$ T.

Report a mistake

Ampere's Law, Cylinders, Solenoid & Toroid question from MHT CET 2019 (3 May, Shift 1)

More questions on Ampere's Law, Cylinders, Solenoid & Toroid