Question

Two long straight parallel conductors carrying steady currents I₁ and I₂ are separated by a distance 'd'. Explain briefly, with the help of a suitable diagram, how the magnetic field due to one conductor acts on the other. Hence deduce the expression for the force acting between the two conductors. Mention the nature of this force.

Answer 1

Assumption: Current flows in the same direction.

Using Right hand thumb rule, the direction of the magnetic field at point P due to current I₂ is perpendicular to the plane of paper and inwards.

Similarly, at point Q on X₂Y₂, the direction of magnetic field due to current I₁ is perpendicularly outward.

Using Fleming’s left hand rule we can find the direction of forces F₁₂ and F₂₁ which are in opposite directions thus,

By Ampere’s circuited law, we have,

`B^2 =mu_0/(4pi) (2I_2)/d`

Now, F₁₂ = I₁LB₂ (Where L ≡ length of the conductors)

`F_12 =(mu_0)/(4pi) (2I_1I_2L)/d =mu_0/(2pi)(I_1I_2L)/d`

In similar manner we get,

`F_21 =mu_0/(2pi) (I_1I_2L)/d  .... (1)`

From above we get the magnitude of forces F₁₂ and F₂₁ are equal but in opposite direction. So,

F₁₂ = −F₂₁

Therefore, two parallel straight conductors carrying current in the same direction attract each other.

Similarly, we can prove if two parallel straight conductors carry currents in opposite direction, they repel each other with the same magnitude as equation (1).