Question

Three long cables running north and south, are tightly enclosed in an insulating sheath. One of the cables carries a 23 A current southward and the other two carry currents of 17 A and 11 A northward. Use Ampere's law to calculate the magnitude of the magnetic field at 10 m from the cables.

Answer 1

Answer:

Explanation:

Magnitude of magnetic field due to a current carrying conductor

= ( μ₀ / 4π ) x 2I / d   , I is current , d is distance from wire . This formula has been derived from Ampere's Law.

= 10⁻⁷ x  2 x 23 / 10 Current given  I = 23

= 46 x 10⁻⁸ T .

Magnetic field due to current 17 A

10⁻⁷ x  2 x 17 / 10

= 34 x 10⁻⁸ T .

Magnetic field due to current 11 A

10⁻⁷ x  2 x 11 / 10

= 22 x 10⁻⁸ T .

magnetic field due to current 17 A and 11 A are in same direction so total magnetic field

= 34 x 10⁻⁸ + 22 x 10⁻⁸

= 56 x 10⁻⁸ T

magnetic field due to current 23 A  in opposite direction so total magnetic field

= 46 x 10⁻⁸ - 56 x 10⁻⁸

= - 10 x 10⁻⁸ T

Magnitude = 10 x 10⁻⁸ T .