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Two parallel horizontal conductors are suspended by light vertical threads 75.0 cm long. Each conductor has a mass of 40.0 gm per metre, and when there is 110 current they are 0.5cm apart. Equal magnitude current in the two wires result in a separatio11 of 1.5cm. Find the values and directions of currents. |
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Answer» Solution :The situation is shown in figure. Here, we have `Tcostheta=mg`  `Tsintheta=F=(mu_(0))/(4PI)l.(2i_(1)i_(2))/d` or `Tsintheta=(mu_(0))/(4pi)l.(2i^(2))/d` from the above equations `tantheta=(mu_(0))/(4pi).1.(2i^(2))/d.(1)/(mg)` where `THETA` is small, `tantheta~~sintheta` From figure `sinteta=(0.5xx10^(-2))/(75xx10^(-2))` `m=40.0xx10^(-3)` 1kg where l = LENGTH of conductor in meter Substituting in eq. (), we get `(0.5xx10^(-2))/(75xx10^(-2))`= `10^(-7).1.(2i^(2))/((1.5xx10^(-2)))xx1/((40xx10^(-3))1xx9.8)` Solving, we get i = 14 amp. As conductors are repelled, the currents in them are in OPPOSITE directions . |
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