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Standard XII

Physics

Current Density Vector

At room tempe...

Question

At room temperature, copper has free electron density of $8.4 \times 10^{28} m^{- 3}$ . The electron drift velocity in a copper conductor of cross-sectional area of $10^{- 6} m^{2}$ and carrying a current of $5.4 A$ , will be

4 {m s}^{- 1}

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0.4 {m s}^{- 1}

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4 c m s^{- 1}

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0.4 m m s^{- 1}

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Solution

The correct option is C $0.4 m m s^{- 1}$
Current density and drift velocity are related by:
$J = n e v_{d} \Rightarrow I = n A e v_{d}$
Substituting the values, we get:
$v_{d} = \frac{I}{n A e} = \frac{5.4}{8.4 \times 10^{28} \times 10^{- 6} \times 1.6 \times 10^{- 19}} = 4 \times 10^{- 4} m / s$

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Similar questions

Q. Copper contains

8.4 \times 10^{28}

free electrons

/ m^{3}

. A copper wire of cross-sectional area

7.4 \times 10^{- 7} m^{2}

carries a current of

1

A. The electron drift speed is approximately.

A copper wire has a cross sectional area of $1 m m^{2}$ . The electron density of copper is $8.5 \times 10^{28} m^{3}$ . Calculate the mean drift velocity of the electrons through the wire when the current is 3 A.

Q. The cross-sectional area of a wire is

1.0 \times 10^{- 7} m^{2}

and the density of free electrons is

2.0 \times 10^{28} m^{- 3}

What will be the drift velocity of the free electrons for a current of

3.2 A

in the wire?

Q. Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area

1.0 \times 10^{- 7} m^{2}

carrying a current of

1.5 A

. Assume the density of conduction electrons to be

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

The number density of free electrons in a copper conductor estimated in Example 3.1 is 8.5 × 10²⁸ m⁻³. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0 × 10⁻⁶m² and it is carrying a current of 3.0 A.

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At room temperature, copper has free electron density of 8.4×1028m−3. The electron drift velocity in a copper conductor of cross-sectional area of 10−6m2 and carrying a current of 5.4 A, will be

The correct option is C 0.4 mms−1Current density and drift velocity are related by:J=nevd⇒I=nAevdSubstituting the values, we get: vd=InAe=5.48.4×1028×10−6×1.6×10−19=4×10−4 m/s

At room temperature, copper has free electron density of $8.4 \times 10^{28} m^{- 3}$ . The electron drift velocity in a copper conductor of cross-sectional area of $10^{- 6} m^{2}$ and carrying a current of $5.4 A$ , will be

The correct option is C $0.4 m m s^{- 1}$
Current density and drift velocity are related by:
$J = n e v_{d} \Rightarrow I = n A e v_{d}$
Substituting the values, we get:
$v_{d} = \frac{I}{n A e} = \frac{5.4}{8.4 \times 10^{28} \times 10^{- 6} \times 1.6 \times 10^{- 19}} = 4 \times 10^{- 4} m / s$