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Two parallel horizontal conductors are suspended by light vertical threads 75.0 cm long. Each conductor has a mass of 40.0gm per metre, and when there is no current they are 0.5 cm a part. Equal magnitude current in the tow wires in a separation of 1.5 cm . Find the values and directions of currents. |
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Answer» Solution :The situation is shown in figure, Here, we have ` T cos theta = mg (i)` `T sin theta = F = (mu_0)/(4 PI) L (2i_1 i_2)/(d) or ` `T sin theta = (mu_0)/(4pi) l (2i^2)/(d)....(ii)` from eqs. (i) and (ii) `tan theta = (mu_0)/(4pi) l . (2i^2)/(d). (1)/(mg)....(iii)`  where `theta` is small, `tan theta ~~ sin theta` From figure `sin theta = (0.5 xx 10^(-2))/(75 xx 10^(-2))` `m = 40.0 xx 10^(-3) l KG` where l = length of conductor in meter Substituting in eq. (iii). we get `(0.5 xx 10^(-2))/(75 xx 10^(-2)) = 10^(-7)l (2i^2)/((1.5 xx 10^(-2))) xx(1)/((40xx 10^(-3))lxx 9.8)` Solving , we get `i = 14` amp. For REPULSION , the currents are in opposite direction. |
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