Question

The Hall effect can be used to calculate the charge-carrier number density in a conductor. A conductor carrying current of 2.0 A is 0.50 mm thick, and the Hall voltage is 4.5 x 10 V when it is in a uniform magnetic field of 1.2 T. What is the density of charge carriers in the conductor? a) n = 4.6 x 1027 charges/m b) n = 1.7 x 1027 charges/m c) n = 1.0 x 1028 charges/m? d) n = 6.7 x 1027 charges/m?

Answer 1

The density of charge carriers in the conductor is
`6.7*10^(27)\ charge/m^3`

(d) is correct option.

Step-by-step explanation:

Given that,

Current = 2.0 A

Thickness = 0.50 mm

Hall voltage
`v= 4.5*10^(-6)\ V`

Magnetic field = 1.2 T

We need to calculate the density of charge carriers in the conductor

Using formula of the density of charge

`V_(h)=(iB)/(neL)`

`n=(iB)/(eVL)`

Where, i = current

B = magnetic filed

V = voltage

L = thickness

Put the value into the formula

`n=(2.0*1.2)/(1.6*10^(-19)*0.50*10^(-3)*4.5*10^(-6))`

`n=6.7*10^(27)\ charge/m^3`

Hence, The density of charge carriers in the conductor is
`6.7*10^(27)\ charge/m^3`