Q. 294.3( 3 Votes )

A current i is passed through a silver strip of width d and area of cross section A. The number of free electrons per unit volume is π.

(a) Find the drift velocity v of the electrons.

(b) If a magnetic field B exists in the region as shown in figure, what is the average magnetic force on the free electrons?

(c) Due to the magnetic force, the free electrons get accumulated on one side of the conductor along its length. This produces a transverse electric field in the conductor which opposes the magnetic force on the electrons. Find the magnitude of the electric field which will stop further accumulation of electrons.

(d) What will be the potential difference developed across the width of the conductor due to the electron-accumulation? The appearance of a transvers emf, when a current carrying wire is placed in a magnetic field, is called Hall effect.



Answer :

Given-
Width of the silver strip = d


Area of cross-section = A


Electric current flowing through the strip = i


The number of free electrons per unit volume = n




(a) We know the relation between the drift velocity of electrons and current through any wire,




where


e is the charge of an electron


vd is the drift velocity.


A is the area of the conductor


On solving for the drift velocity, we get



(b)The magnetic field existing in the region is B


.
Magnetic force acting due to presence of current on a small length l given by –




where,


= magnetic field


I = current


= length of the wire


So, the force on a free electron





which acts towards upward direction


(c) Let us consider, electric field as E.


Now, the accumulation of electrons will stop when magnetic force just balances the electric force.


Electric force, coulomb’s law,



where


e = charge


E=electric field of the charge


From (1) and (2)




(d)The potential difference , V developed across the width d of


the conductor due to the electron-accumulation is given by-



where


d is length of wire


and E is applied electric field



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