A current flows to the right through a rectangular bar of conducting material, in the presence of a uniform magnetic field$\mathit{B}$pointing out of the page (Fig. 5.56).

(a) If the moving charges are positive, in which direction are they deflected by the magnetic field? This deflection results in an accumulation of charge on the upper and lower surfaces of the bar, which in turn produces an electric force to counteract the magnetic one. Equilibrium occurs when the two exactly cancel. (This phenomenon is known as the Hall effect.)

(b) Find the resulting potential difference (the Hall voltage) between the top and bottom of the bar, in terms of$\mathit{B}$,$\mathit{v}$(the speed of the charges), and the relevant dimensions of the bar.${}^{\mathbf{23}}$

(c) How would your analysis change if the moving charges were negative? [The Hall effect is the classic way of determining the sign of the mobile charge carriers in a material.]

The given data is listed below as:

• The conducting material has a current $I$.
• The uniform magnetic field of the conducting bar is,$B$
• The width of the bar is,$w$
• The length of the bar is,$I$
• The thickness of the bar is,$t$

The magnetic field is described as a particular region that helps a magnetic material to exert a magnetic force on other objects. Moreover, this magnetic field

is also termed as a vector field that is beneficial for describing the magnetic influence on the moving charges.

According to the hall effect, if the positive charges mainly flow to the right, then the bottom plate also acquires a positive charge. Hence, due to flowing in the right direction, the positive charges are also deflected downwards.

Thus, the positive charges are deflected in the downwards direction by the magnetic field.

The equation of the force at the top of the bar is expressed as:

$F=qvB$ …(i)

Here,$q$is the electric charge,$v$is the velocity of the bat and$B$is the uniform magnetic field.

The equation of the force at the bottom of the bar is expressed as:

$F=qE$ …(i)

Here, $q$is the electric charge and$E$is the electric field.

As the force on the top and the bottom of the bar is equal. Then the equation (i) and (ii) is also equal.

Equating the equation (i) and (ii).

$$\begin{gathered} qvB=qE \\ E=vB \end{gathered}$$ …(iii)

The equation of the potential difference between the top and the bottom of the bar is expressed as:

$V_1-V_2=Et$

Here,$V_1$is the potential at the top and$V_2$is the potential at the bottom of the bar and$t$is the thickness of the bar.

Substitute $vB$for $E$from the equation (iii) in the above equation.

$V_1-V_2=vBt$

Thus, the resulting potential difference between the top and the bottom of the bar is$vBt$ .

According to the Hall effect, if the negative charges mainly flow to the left, then the negative charge is also deflected downwards and the bottom plate becomes negatively charged. Moreover, the difference in potential between the two plates is the same but the top plate mainly acquires a higher potential.

Thus, the negative charge is deflected downwards and the top plate gets the higher potential.