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Standard XII

Physics

Electric Field Due to Charge Distributions - Approach

There is a cu...

Question

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is $1 {mm}^{2}$ . If the number of free electrons per ${cm}^{3}$ is $8.4 \times 10^{22}$ , then the drift velocity of electrons will be

1.0 mm per sec

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1.0 meter per sec

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0.1 mm per sec

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0.01 mm per sec

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Solution

The correct option is C 0.1 mm per sec
$I = n e A v_{d} \Rightarrow 1.344 = 8.4 \times 10^{22} \times 10^{6} \times 1.6 \times 1.6 \times 10^{- 19} \times 1 \times 10^{- 6} \times V_{d}$
$∴ v_{d} = 1 \times 10^{- 4} m/s = 0.1 mm / s$

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Similar questions

Q. There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is

1 m m^{2}

. If the number of free electrons per

c m^{3} i s 8.4 \times 10^{22}

, then the drift velocity would be

Q. There are

8.4 \times 10^{22}

free electrons per

c m^{3}

in copper, the current in the wire is

0.21 A

(

e = 1.6 \times 10^{- 19} C

). Then the drifts velocity of electrons in a copper wire of

1 m m^{2}

cross section will be:-

Q. There are

8.4 \times 10^{22}

free electrons per

{cm}^{3}

in copper. The current in the wire is

0.21

A

(e = 1.6 \times 10^{- 19} C

). Then the drift velocity of electrons in a copper wire of

1

{mm}^{2}

cross section will be

Q. The current of

0.5

Amp is flowing in a copper wire whose cross-sectional area is

1 m m^{2}

. If the number of free electrons is

8.5 \times 10^{22} p e r c m^{3}

in unit volume then find out the drift velocity of electrons?

Q. A uniform wire of length

2.0 m

and cross-sectional area

10^{- 7} m^{2}

carries a current of

1.6 A

. If there are

10^{28}

free electrons per

m^{3}

in copper, the drift speed of electrons in copper is

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There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is 1 mm2. If the number of free electrons per cm3 is 8.4×1022, then the drift velocity of electrons will be

The correct option is C 0.1 mm per secI=neAvd⇒1.344=8.4×1022×106×1.6×1.6×10−19×1×10−6×Vd ∴vd=1×10−4 m/s=0.1 mm/s

There is a current of 1.344 amp in a copper wire whose area of cross-section normal to the length of the wire is $1 {mm}^{2}$ . If the number of free electrons per ${cm}^{3}$ is $8.4 \times 10^{22}$ , then the drift velocity of electrons will be

The correct option is C 0.1 mm per sec
$I = n e A v_{d} \Rightarrow 1.344 = 8.4 \times 10^{22} \times 10^{6} \times 1.6 \times 1.6 \times 10^{- 19} \times 1 \times 10^{- 6} \times V_{d}$
$∴ v_{d} = 1 \times 10^{- 4} m/s = 0.1 mm / s$