Question

A current i, indicated by the crosses in figure, is established in a strip of copper of height h and width w. A uniform field of magnetic induction B is applied at right angle to the strip. If no electric field is applied from the outside, the electrons will be pushed somewhat to one side and therefore will give rise to a uniform electric field $E_H$ across the conductor until the forces of this electrostatic field $E_H$ balance the magnetic forces encountered in part (b). What will be the magnitude and direction of field $E_H$? Assume that n, the number of conductor electrons per unit volume is $1.1\times {10}^{29}{m}^{-3},h=0.02m,w=0.1cm,i=50A$, and $B=2T$

• A. $2.8\times {10}^{-4}$ , downward
• B. $2.8\times {10}^{-4}$ , upward
• C. $1.2\times {10}^{-4}$ , downward
• D. $1.2\times {10}^{-4}$ , upward

Answer 1

The correct option is A $2.8\times {10}^{-4}$ , downward

As direction of motion of electron is opposite to that of direction of current so electron moves out from the plane
Due to magnetic field, a force will act on $e^{-}$. Direction of force is in downward direction
$\overrightarrow{F_m}=-e(\overrightarrow{V_d}\times \overrightarrow{B})$
This will accumulate electrons in bottom and $e^{-}$ deficiency at top. Due to this electric field is generated in downward direction. More $e^{-}$ accumulate the intensity of E increases and so due to electric field on incoming electron increases. Electrostatic force is in upward direction
$\overrightarrow{F_E}=-e\overrightarrow{E}$
At equilibrium condition
Magnetic force on incoming $e^{-}$=electrostatic force on incoming $e^{-}$
$\Rightarrow -e{V}_{d}B\sin 90=-eE$
$\Rightarrow E=V_{d}B$
where $V_d\to$drift velocity $;A\to$cross sectional area
Now current $i=neA{V}_d$
$E=\Rightarrow 50\Rightarrow 1.1\times {10}^{29}\times 1.6\times {10}^{-19}\times (w\times h)\times V_d$
$V_d=\frac{50}{1.6\times 1.1\times {10}^{10}\times 1\times {10}^{-3}\times 2\times {10}^{-2}}$
$=\frac{25}{1.6\times 1.1}\times {10}^{-5}$
$=14.2045\times {10}^{-5}=1.42045\times {10}^{-4}㎧$
$E=V_{d}B=1.42045\times {10}^{-4}\times 2$
$E=2.8\times {10}^{-4}N/C$downward