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This mock test of Test: The Hall Effect for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
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QUESTION: 1

In the Hall Effect, the directions of electric field and magnetic field are parallel to each other.

The above statement is

Solution:

To make Lorentz force into the effect, the electric field and magnetic field should be perpendicular to each other.

QUESTION: 2

Which of the following parameters can’t be found with Hall Effect?

Solution:

The Hall Effect is used for finding the whether the semiconductor is of n-type or p-type, mobility, conductivity and the carrier concentration.

QUESTION: 3

In the Hall Effect, the electric field is in x direction and the velocity is in y direction. What is the direction of the magnetic field?

Solution:

The Hall Effect satisfies the Lorentz’s Force which is

E=vxB

So, the direction of the velocity, electric field and magnetic field should be perpendicular to each other.

QUESTION: 4

What is the velocity when the electric field is 5V/m and the magnetic field is 5A/m?

Solution:

E=vxB

v=E/B

=5/5

=1m/s.

QUESTION: 5

Calculate the hall voltage when the Electric Field is 5V/m and height of the semiconductor is 2cm.

Solution:

Vh=E*d

=5*2/100

=0.1V.

QUESTION: 6

Which of the following formulae doesn’t account for correct expression for J?

Solution:

B=µH

So, option d is correct option.

QUESTION: 7

Calculate the Hall voltage when B=5A/m, I=2A, w=5cm and n=10^{20}.

Solution:

Vh=BI/wρ

=5=2/ (5*10^{-2}*10^{5}*1.6*10^{-19})

=0.002V.

QUESTION: 8

Calculate the Hall Effect coefficient when number of electrons in a semiconductor is 10^{20}.

Solution:

R=1/ρ

=1/(1.6*10^{-19}*10^{20})

=0.0625.

QUESTION: 9

What is the conductivity when the Hall Effect coefficient is 5 and mobility is 5cm^{2} /s.

Solution:

µ=σR

σ =µ/R

=5*10^{-4}/5

=0.0001 S/m.

QUESTION: 10

In Hall Effect, the electric field applied is perpendicular to both current and magnetic field?

Solution:

In Hall Effect, the electric field is perpendicular to both current and magnetic field so that the force due to magnetic field can be balanced by the electric field or vice versa.

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