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Two long thin parallelconductors of the shapeshown in Fig. carry direct currentsI_(1) and widthof the right-handconductoris equal to b. With bothconductorslying in one plane, find themagnetic interactionforce betweenthem reducedto a unitof their length. |
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Answer» SOLUTION :We know that Ampere'sforce per unit length on a wireelementin a magnetic field isgiven by `d vec(F_(n)) = I (hat(n) xx hat(B))` where `hat(n)` is the unit vector alongthe directionof current...(1) Now, let us take an element OFTHE conductor`i_(2)` as shwon in the fig. The wireelementis in the magnetic field, PRODUCEDBY the current`i_(1)`which is directednormallyinto the sheetof the paperand itsmagnitudeis given by, `|vec(B)| = (mu_(0) l_(1))/(2pi r)` ....(2) From Eqs. (1) and (2) `d vec(F_(n)) = (I_(2))/(b) dr (hat(n) xx vec(B))`. (BECUASE the currentthroughthe element equals `(l_(2))/(b) dr`) So, `d vec(F_(n)) = (mu_(0))/(2pi) (l_(1) l_(2))/(b) (dr)/(r)`, towardsleft (as `hat(n) _|_^(r) vec(B)`) HENCETHE magnetic force on the conductor: `vec(F_(n)) = (mu_(0))/(2pi) (I_(1) I_(2))/(b) int_(a)^(a + b) (dr)/(r)` (towards left) `= (mu_(0) I_(1) I_(2))/(2pi b) ln (a + b)/(a)` 
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