Question

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are l₁ = + 10 A, l₂ = -13 A, l₃ = + 10 A, l₄ = + 7 A, l₅ = -12 A and l₆ = + 18 A. The magnetic induction at a perpendicular distance of 10 cm from the cable is ______.
(µ₀ = 4π x 10^-7 Wb/A-m)

Options

• 40 µT

• 37.5 µT

• 30 µT

• 35 µT

Answer 1

Six very long insulated copper wires are bound together to form a cable. The currents carried by the wires are l₁ = + 10 A, l₂ = -13 A, l₃ = + 10 A, l₄ = + 7 A, l₅ = -12 A and l₆ = + 18 A. The magnetic induction at a perpendicular distance of 10 cm from the cable is 40 µT.

Explanation:

Net current due to all wires.

`"i"_"net" = "i"_1 + "i"_2 + "i"_3 + "i"_4 + "i"_5 + "i"_6`

`"i"_"net"` = 10 - 13 + 10 + 7 - 12 + 18 = 20 A

We know, magnetic field due to an infinitely long straight conductor at a perpendicular distance r from it is given by

B = `(mu_0"i")/(2pi"r") = (mu_0 "i"_"net")/(2pi"r")`

where, i = current in wire

and r = perpendicular distance.

`= (4pi xx 10^-7 xx 20)/(2pi xx 10 xx 10^-2) = 4 xx 10^-5`

B = 40 µT