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Standard XII

Physics

Drift Velocity

A current of ...

Question

A current of $1 A$ flows through a copper wire. The number of electrons passing through any cross-section of the wire in $1.6 s e c$ is (charge of a electrons $= 1.6 \times 10^{- 19} C$ )

10^{19}

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10^{22}

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10^{25}

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10^{28}

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Solution

The correct option is A $10^{19}$
Current $= \frac{charge of electrons following}{time taken to flow}$
$I = \frac{n (1.6 \times 10^{- 19})}{1.6}$
$n = \frac{1.6 \times 10^{19}}{1.6} = 10^{19}$ electrons

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Similar questions

Q. When a steady current of

1 A

flows through a copper wire, the number of electrons crossing any cross-section area of the wire in one second is ______.

The charge of an electron is known to be $1.6 \times 10^{- 19} C$ . In a circuit, the current flowing is $1 A$ . How many electrons will be flowing through the circuit per second?

Q. Find the current flows through a copper wire of length 0.2 m, area of cross-section

1 m m^{2}

, when connected to a battery of 4V. Given that electron mobility is

4.5 \times 10^{-} 6 m^{2} s^{-} 1

and charge on an electron is

1.6 \times 10^{- 19} C

The number density of electron in copper wire is

8.5 \times 10^{2} 8 m^{3}

Q. Drift speed of electrons, when

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5 m m 2

, is

v

. If the electron density in copper is

9 \times 10^{28} / m^{3}

the value of

v

m m / s

is close to (Take charge of electron to be

= 1.6 \times 10^{- 19} C

)

Q. If

10

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electrons pass through the cross-section of a wire in

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A current of 1A flows through a copper wire. The number of electrons passing through any cross-section of the wire in 1.6sec is (charge of a electrons =1.6×10−19C)

The correct option is A 1019Current =charge of electrons following time taken to flowI=n(1.6×10−19)1.6n=1.6×10191.6=1019 electrons

A current of $1 A$ flows through a copper wire. The number of electrons passing through any cross-section of the wire in $1.6 s e c$ is (charge of a electrons $= 1.6 \times 10^{- 19} C$ )

The correct option is A $10^{19}$
Current $= \frac{charge of electrons following}{time taken to flow}$
$I = \frac{n (1.6 \times 10^{- 19})}{1.6}$
$n = \frac{1.6 \times 10^{19}}{1.6} = 10^{19}$ electrons