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Question: Suppose a two dimensional rectangular material carries a current of \( 0.5A \) in the positive x-dir...

Suppose a two dimensional rectangular material carries a current of 0.5A0.5A in the positive x-direction and is penetrated by a magnetic field of 1.4mT1.4mT in the negative z-direction. The density of mobile charges per unit area of material is 0.2μCm20.2\mu C{m^{ - 2}} , what is the magnitude of the Hall voltage and the direction of the generated electric field? (Assume a right handed coordinate system)
(A) 70μV70\mu V , negative y-direction
(B) 70μV70\mu V , positive y-direction
(C) 3.5kV3.5kV , negative y-direction
(D) 3.5kV3.5kV , positive y-direction

Explanation

Solution

Hint : Here, we have been asked about the Hall voltage and it is calculated by using the principle of Hall Effect. Hall Effect means when a conductor or semiconductor with current flowing in one direction was introduced perpendicular to a magnetic field a voltage could be measured at right angles to the current path. Let us apply this concept and find voltage.

Complete Step By Step Answer:
according to the given information we have
I=0.5AI = 0.5A , B=1.4mTB = 1.4mT and NeA=0.2μCm2\dfrac{{Ne}}{A} = 0.2\mu C{m^{ - 2}}
Formula for the Hall voltage is given by
VH=IBned{V_H} = \dfrac{{IB}}{{ned}}(1)(1)
Where,
II is the current flowing through the conductor.
BB is the magnetic field of the conductor.
ee is the charge.
nn is the number of charge carriers per unit volume.
dd is the thickness of the conductor.
Also, we have to use NeA=nde=0.2μCm2\dfrac{{Ne}}{A} = nde = 0.2\mu C{m^{ - 2}}
In eq(1)eq(1) we have to put all the given data as follows:
eq(1)VH=(0.5)(1.4×103)(0.2×106)eq(1) \Rightarrow {V_H} = \dfrac{{(0.5)(1.4 \times {{10}^{ - 3}})}}{{(0.2 \times {{10}^{ - 6}})}}
VH=3.5kV\Rightarrow {V_H} = 3.5kV
So, we have calculated the Hall voltage now we have to find the direction of the generated electric field.
Let us consider the right handed coordinate system as below:

So, negative y-direction as IB{I_B} and (I×B)l\left( {\vec I \times \vec B} \right)l is in positive y-direction.
Thus, FE=FB={\vec F_E} = - {\vec F_B} = negative y-direction.
Thus, the answer is that the Hall voltage is 3.5kV3.5kV and the electric field is in negative y-direction.

Note :
We must study the facts of Hall Effect and how it works to find hall voltage. We have to use the right handed coordinate system. Thus we have concluded above that the voltage is 3.5kV3.5kV and the electric field is in negative y-direction.