Question

Please show work and explanation. thanks

Physics 51 Practice Final Problem #1 : Two electrons are at opposite corners of a 10cm-per-side square, as shown. A long wire carrying 200mA runs through another corner of the square. The bottom electron is moving towards the top electron at lkm/s electron 10cmV-lkm/s I-200mA nto page electron 10cm A) Find the magnitude and direction of the electric and magnetic fields at the location of the moving electron (the bottom one). Make sure your direction is well-defined (if you use ij,k or x.y,z vectors then "define them) B) Find the magnitude and direction of the TOTAL force on the moving electron (the bottom one). Make sure your direction is well-defined

Answer 1

note:total force is the sum of magnetic and electric force.

The electric field due to electron at the top inner E = 1/4 pi E_0 e/r^2 = (I + J/squareroot 2) r^2 = 10^2 + 10^2 = 200 = -9 times 10^9 times 1.6 times 10^-19/alpha squareroot 2 (I + J) = 10.18 times 10^-10 (I + J)V Magnetic field due to the current carrying wire at distance of 10m B = mu_0 I/2 pi S = 4 pi times 10^-7 times 200 times 10^-3/2 pi times 0.1 = 4000 times 10^-10 T (along y- direction an be found using right hand thumb rule by placing right hand around the current carrying wire) B) Lorentz force law F = qE + qV times B = eE + e VB sin theta here theta = 45 degree therefore F = 1.6 times 10^-19 times 10.18 times 10^-10 + 1.6 times 10^-19 times 1000 times 4000 times 10^-10 = 1/squareroot 2 force due to magnetic field = 6.4 times 10^-23 N towards the reader (since cross product of V and B gives their direction) force due to electric field