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(a) Derive an expression for the force between two long parallel current carrying conductors . (b) Use this expression to define S.I. unit of current . (c) A long straight wire AB carries a current I.A proton P travels with a speed v , parallel to the wire , at a distance d from it in a direction opposite to the current as shown in the figure . What is the force experienced by the proton and what is its direction ? |
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Answer» Solution :(c) Magnetic field induction at the POINT P is `B= (mu_(0))/(4 pi) . (2I)/(r)` The direction of `vecB` is perpendicular to the plane of paper directed inwards (ACCORDING to Right Hand Thumb rule ) The force on moving proton of charge q due to magnetic filed B is F = qvB sin `90^(@) = q V B ` Since charge of proton q = `1.6 XX 10^(-19) C` `therefore F = 1.6 xx 10^(-9) xx v B N.` The direction of force on proton , according to FLEMMING's left Hand Rule acts in the plane of paper towards right . |
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