InterviewSolution
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InterviewSolution
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Two long straight parallel current conductors are kept at a distance d from each other in air. The direction of current in both the conductors is the same. Find the magnitude and direction of the force between them. Hence, define one ampere. |
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Answer» Solution :Consider two straight parallel long current carrying conductors AB and CD carrying currents, `I_1 and I_2` RESPECTIVELY in same direction and let these be separated by a distance d. Now magnitude field `B_1` developed at a point Q on 2nd conductor due to current `I_1` flowing in 1st conductur is `B_1 = (mu_0 I_1)/(2 pi d)` As per right hand rule `B_2` is acting normal to the plane of the paper pointing inward. Thus, conductor CD carrying current `I_2` is in a MAGNETIC field which is perpendicular to its length. Therefore, force experienced by 2nd conductor CD due to `B_1` `F_(21) = B_1 I_2, l,` Where l = length of the 2nd condcutor or `F_(21) = (mu_0 I_1)/(2 pi d) I_2 l = (mu_0 I_1 I_2 l)/(2 pi d)` and force per unit length `(F_(21))/(l) = (mu_0 I_1 I_2)/(2 pi d) = (mu_0)/(4pi) CDOT (2 I_1 I_2)/(d)` The force `F_(21)` in accordance with Fleming.s LEFT hand rule is directed towards the conductor AB. In the same way, it is FOUND that force experienced per unit length of wire AB is `(F_(21))/(l) = (mu_0)/(4pi) cdot (2 I_1 I_2)/(d)` and is directed towards CD. 
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