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When 5V potential difference is applied ...

When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electron a `2.5 times 10^-4 m.s^-1`. IF the electron density in the wire is `8 times 10^28 m^-3` , the resistivity of the material is close to

A

`1.6 times 10^-8 Omega.m.`

B

`1.6 times 10^-7 Omega.m`

C

`1.6 times 10^-6 Omega.m.`

D

`1.6 times 10^-5 Omega.m`

Text Solution

Verified by Experts

The correct Answer is:
D
Drift velocity, `v=I/(n e A)`
`I=V/R=V/(rho1/A)or,rho=V/(l/Al)=V/(n e vl)`
`thereforerho=5/((8times10^28)times(1.6times10^-19)times(2.5 times 10^-4)times0.1)`
`1.56 times 10^-5Omega.m.=1.6 times 10^-5 Omega.m.`
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Knowledge Check

  • When 5V potential difference is applied across a wire of length 0.1 m, the drift speed of electrons is 2.5 xx I0^(-4)ms^(-1) . If the electron density in the wire is 8xx10^28m^(-3) the resistivity of the material! is close to

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  • If a current passes through a metal conducting wire of area of cross section A, the drift velocity of free electrons inside the metal is v_d=1/( n e A) , where the amount of electric charge of an electron =e and the number of free electrons per unit volume of the metal=n. The applied electric field on the wire is E=V/l , where a potential difference V exists between two points, l apart, along the length of the wire. IF R is the resistance of the wire between those two points, then the resistivity of its material is rho=(RA)/l .Besides the mobility (mu) of the free electrons inside a wire is defined as their drift velocity for a unit applied electric field. The radii of two wires ,made of two different metals are in the ratio 1:2 , The number density of free electrons in the first metal is double that in the second metal. IF the current in the first wire is 1A, then the current in the second wire producing the same drift velocity is

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  • If a current passes through a metal conducting wire of area of cross section A, the drift velocity of free electrons inside the metal is v_d=1/( n e A) , where the amount of electric charge of an electron =e and the number of free electrons per unit volume of the metal=n. The applied electric field on the wire is E=V/l , where a potential difference V exists between two points, l apart, along the length of the wire. IF R is the resistance of the wire between those two points, then the resistivity of its material is rho=(RA)/l .Besides the mobility (mu) of the free electrons inside a wire is defined as their drift velocity for a unit applied electric field. The radii of two wires of the same metal are in the ratio 1:2 The same potential difference is applied between two points at a distance l on each of the wires.The ratio between the drift velocities of the free electrons in two wires is

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    `1:1`
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    C
    `2:1`
    D
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