Answer :
Option (d) is correct, According to the given conditions here we apply the right hand rule, the force pushing on an electron is travelling in the x direction.
The width of sheet is: wo=4.5cmwo=4.5cm, The current density is: σo=5200A/m2σo=5200A/m2, The mass of electron is: me=9.11×10−31kgme=9.11×10−31kg, The charge of electron is: qe=−1.60×10−19Cqe=−1.60×10−19C, The velocity of electron is: Ve=2.5×106m/sVe=2.5×106m/s, The initial position of electron is: z = 4.0cmz = 4.0cm
Io=oAo(I) The expression for the current produced by the sheet is Io=oAo. (I)
Ao=wozAo=wozis the formula for sheet area.
Change the value in the previous expression.
Ao=0.045m×0.04mAo=0.0018m2
Ao=0.045m×0.04mAo=0.0018m2
It is necessary to adjust the value in the expression (I) to Io=5200A/m20.0018m2Io=9.36A.
The magnetic field equation is Bo=sIo2z. Bo=μsIo2πz
The term "permeability of open space" (ss) is used here.
The value of an open space's permeability is as follows: s=4107T.m/A
Change the value in the previous expression.
Bo=4π×10−7T.m/A×9.36A2×π×(0.04m)
Bo=0.0000468TBo=4π×10−7T.m/A×9.36A2×π×(0.04m)
Bo=0.0000468T
The force on an electron has the formula Fo=qeBoVe.
Fo=qeBoVe
Change the value in the previous expression.
Fo=−1.60×10−19C×2.5×106m/s×0.0000468TFo=−1.872×10−17NFo=−1.60×10−19C×2.5×106m/s×0.0000468TFo=−1.872×10−17N
According to the right hand rule, the force pushing on an electron is travelling in the x direction.
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