Question

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 n. (a) Find the drift velocity v of the electrons. (b) If a magnetic field B exists in the region, as shown in the 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 transverse emf, when a current-carrying wire is placed in a magnetic field, is called Hall effect.
Figure

Answer 1

Explanation:

It is given:

Silver strips’s width = d

Cross-section’s area = A

Electric current = i

Per unit volume, the number of free electrons = n

(a) The relation between the current and drift velocity,

i = vnAe, where v is the drift velocity and e = charge of an electron.

v = inAe

(b)The magnetic field existing.

On a current-carrying conductor, the average magnetic force,

F = ilB

So, a free electron’s force = ilBnAl = iBnA upwards  

(c) Consider the electric field as E.

Thus, accumulating the electrons further will halt when there is this balance of electric force by the magnetic force.

So, E = iBeAn

(d) The potential difference advanced across the conductor’s width because of the accumulation of electron will be V = Er.