Question

magnetic flux
As shown in the figure, there are semiconductors with a cross-section of width w [m] and thickness t [m].
I [A] current is flowing in this semiconductor. However, the electron density per unit volume is N [m-3].
When a magnetic field of magnetic flux density B [T] exists in this space in the direction shown in the figure, an electromotive force is generated in the direction of the semiconductor. Find its direction and size

В W

Answer 1

A =wt mo Plared Pn Q. Given that Semiconductor of poss-sectional width to and thickness t The cross-sectional Area will be The semiconductor hos cament and mognetic field B ( Both are perpendicular the magnetic field B & B f and let Va is drift velocity of charge to each other) ny B KEM + + + AR Z 48 А The canent density is given by J neva Vy = emp where number density of elecbons and e = charge of .elechon, The cusent density J = canent Area (cross sectional) Vo O mogretic • The force e lechor for electon 9-e and J = Va fîl opposite to current The magnetic force F F = q B x B = -e va B (2x9) on electron Em = evdB (î •-| = neud ne A is on

Due toi magnetic force on elechon towards -X axis and similorly the Mognetic force on hole will be towards tx axis these forces deflects the charges along + axis (hole ) and - axrs (eleebon) Due to which a elechic field created along the titek axis ( tue to the (Ex) Charge ) on charge a q va B qEH Due to this electric field , electric force acts (Fe) At eguillibrum Fm = (mognetic force on charge equal to electric force on charge) Vd B = En also En Creates a Potential difference (VH) emf across the width which is given by so VH VH EH W EH VH Ve B Va 1 hen he A Vd น WB VH WB P VH ľ Bo ne A PBw he wt he A VH iB along width net The emf genrated . VA = iB net wong cand the (The electric field & towords negative X axis )