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Two long straight parallel conductors carrying currents I_1 and I_2 along the same direction are separated by a distance 'd'. How does one explain the force of attraction between them? If a third conductors carrying a current I_3 in the opposite direction is placed just in the middle of these conductors , find the resultant force acting on the third conductor. |
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Answer» Solution :Let two long straight parallel conductor carrying currents `I_1 and I_2` in same direction are separated by a DISTANCE .d. (fig.). Then at a point N on conductor 2, a magnetic field `B_1 = (mu_0 I_1)/(2 pi d)` is set up due to `I_1`, and it is directed normal to hte plane of paper of paper poiniting into it. The conductor 2 caryying CURRENT `I_2`experience a force per unit length `F_(21) = B_(1) I_2 = (mu_0 I_1 I_2)/(2 pi d)`, whose direction in accordance with Fleming.s left hand rule is toward conductor 1. Thus, the force is attractive in nature. Let a conductor 3 carrying current `I_3` in opposite direction be placed just in the MIDDLE of these conductors. Then this conductor experiences force `vec(F_(31))` due to condcutor 1 and `vec(F_(32))` due to conductor 2, which are in the directions as shown in Fig. Obivously net force `vec(F_3) = vec(F_31) - vec(F_32) = (mu_0 I_1 I_3)/(2 pi (d/2)) - (mu_0 I_2 I_3)/(2 pi (d/2))` `= (mu_0 I_3)/(pi d) (I_1 - I_2)` towards conductor 2. 
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