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Give expression for the force of a current carrying conductor in a magnetic field. |
Answer» SOLUTION :1. As shown in figure, consider a conductor PQ of length l are of cross-section A, carrying current I along + ve y-direction. The field `vecB` ACTS along + ve z-direction.  2. The electrons drift towards left with velocity `vecv_(d)`. 3. Each electron experience force along + ve X-axis which is given by, `vecf=-e(vecv_(d)xxvecB)` 4. If n is the number of free electrons per unit volume, then total number of electrons in the conductor is, `N=nxx"Volume"=nAl` 5. Total force on the conductor is, `vecF=Nvecf=nAl[-e(vecv_(d)xxvecB)]=nAe(-(lvecv_(d)xxvecB))` But `Ivecl` represents a current element vector in the direction of current so we can take `-lvecv_(d)=v_(d)vecl` `thereforevecF=nAe(v_(d)veclxxvecB)=nAev_(d)(veclxxvecB)` but `nAev_(d)` = current I `thereforevecF=I(veclxxvecB)` and magnitude F = `IlBsintheta` where `theta` is angle between `vecBandI`. 6. This equation can be applicable for straight conducting rod. 7. If the wire has an arbitrary shape we can calculate the Lorentz force on it by considering it as a collection of linear strips dl and summing, `thereforevecF=sum_(i=1)^(n)vec(dl)_(i)xxvecB` where, i = 1, 2, 3, ...., n 8. Direction of this force can be given by Flemming.s left hand rule. 9. Special Cases : (i) If `theta=0^(@)or180^(@)` then, `F=IlBsin0^(@)=0""[becausesin0^(@)=0andsin180^(@)=0]` Thus, a current carrying conductor placed PARALLEL to the direction of the MAGNETIC field does not experience any force. (ii) If `theta=90^(@)`, then `F=IlBsin90^(@)` `thereforeF=IlB""[becausesin90^(@)=1]` Thus, a current carrying conductor placed perpendicular to the direction of a magnetic field experience a maximum force. |
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