CSEM02 20.HW.16.alt Pull a rectangular metal loop through a magnetic field(a,b) The mobile electrons in the portion of the loop that is exposed to the magnetic field experience amagnetic force. This magnetic force causes the bar (vertical segment of the loop) to polarize, as shown below.The polarization continues until the electric force on an electron in the moving bar is equal and opposite tothe magnetic force on the electron.(c) The round-trip potential difference in the loop is zero:The emf (due to the magnetic force on the mobile charges) is numerically equal to the potential differenceacross the polarized segment of the loop:vBvBFmagFelvBFmagvBFmagFelvhE∆V = EhIvhE∆V = EhIBar moving to right;magnetic force causeselectrons to move down.When polarization creates alarge enough electric field inthe bar, the net force in thevertical direction becomeszero.The polarized end of the loop actslike a battery, causing current toflow through the loop. In this caseconventional current flows clock-wise.Bar moving to left; mag-netic force causes elec-trons to move up.If the loop is moving to the right:If the bar is moving to the left:When polarization creates alarge enough electric field inthe bar, the net force in thevertical direction becomeszero.The polarized end of the loop actslike a battery, causing current toflow through the loop. In this caseconventional current flows coutner-clockwise.∆VRT0emfIR–==IemfR--------- -=emf Eh=Inside the bar, the vertical component of the net force on a mobile electron is almost zero in the steady state(it is not exactly zero, since as long as current is flowing, electrons must still move toward the negative end ofthe bar to replace those that flow away; but the net vertical force on a mobile electron is very small). So:So the conventional current in the bar is: Using these numerical values: , , , and , we get:(d,e) Now that a current is running through the loop, in the sections of the loop inside the region wherethere is a magnetic field, there is an additional force on the moving electrons due to their drift velocity. Theforces on the top and bottom segments of the loop cancel (though they do stretch or compress the loop).However, on the vertical segment of the loop, the magnetic force acts in a direction to slow the loop down.This makes sense energetically; it is necessary to put energy into the system (by doing work in pulling theloop) in order to get the current to flow.Since we have already derived an expression for the magnetic force on a segment of a wire carrying a con-ventional current, we can use this expression to calculate the magnetic force on the vertical segment:Using these numerical values: , , , and , we get: opposite to the loop’s motionSo you must apply a force of the same magnitude in the direction of the loop’s motion, to keep it moving atconstant velocity.FelFmag=eE ev B×=EvB90°si n vB==emf Eh vBh==IvBhR--------- -=h 3 cm= R 0.3 Ω= v 8 m/s= B 1.2 T=I8 m/s()1.2 T()0.03 m()0.3 Ω()-------------------------------------------------------------- 0.96 A==vhIFmag on leftsidevhIFmag on left sideIf the loop is moving to the right:If the loop is moving to the left:Fmag on loopI∆LB× I∆LB 90°sin I∆LB and since here ∆Lh== = =Fmag on loopvBhR--------- -⎝⎠⎛⎞hBvB2h2R-------------- -==h 3 cm= R 0.3 Ω= v 8 m/s= B 1.2 T=Fmag on loop8 m/s()1.2 T()20.03 m()20.3 Ω()------------------------------------------------------------------- 0.0345