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Derive the expression for the magnetic force experienced by a current carrying conductor . |
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Answer» Solution :Consider a CONDUCTING rod of LENGTH l, and uniform sectional area A in an external magnetic field `vecB`. Let the number densityof the mobilecharge carriers (electrons) be n. `therefore` Number density (n) =`"no. of electrons"/"Volume"` no. of electrons = n x volume no. of electrons = n x Al Ifq is the charge of each charge CARRIER Total charge = q x nAl When a steady CURRENT I flows in this conducting rod , each mobile charge carrier has an average drift velocity`vecV_d` in the presence of an external magnetic field `vecB`. `therefore` Force on these charge carriers is `vecF` =Total Charge `(vecVxxvecB)` `vecF=qnAl(vecV_d xx vecB)` `vecF=Al(vecjxxvecB)` `vecF=I(veclxxvecB)` Magnitude `F=BI l SIN theta` Where `vecl` is a vector of magnitude l and vector sign is transferred from j to `vecl` direction of `vecl` is same as j (Current ) . |
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