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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.
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Solution
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).
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