AP EAMCET · PHYSICS · Current Electricity

An electron takes $40 \times 10^{3} s$ to drift from one end of a metal wire of length $2 m$ to its other end. The area of cross-section of the wire is $4 {mm}^{2}$ and it is carrying a current of $1.6 A$ . The number density of free electrons in the metal wire is

A $8 \times 10^{28} m^{- 3}$
B $6 \times 10^{28} m^{- 3}$
C $4 \times 10^{28} m^{- 3}$
D $5 \times 10^{28} m^{- 3}$

Verified Solution

Answer & Solution

Correct Answer

(D) $5 \times 10^{28} m^{- 3}$

Step-by-step Solution

Detailed explanation

From the given data, drift velocity can be written as, v=2 m40×103 s=12×104 m s-1 Now for current we can write, I=neAv⇒n=IeAv⇒n=1.61.6×10-194×10-612×104=5×1028 m-3

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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An electron takes 40×103 s to drift from one end of a metal wire of length 2 m to its other end. The area of cross-section of the wire is 4 mm2 and it is carrying a current of 1.6 A. The number density of free electrons in the metal wire is

Step-by-step Solution

An electron takes $40 \times 10^{3} s$ to drift from one end of a metal wire of length $2 m$ to its other end. The area of cross-section of the wire is $4 {mm}^{2}$ and it is carrying a current of $1.6 A$ . The number density of free electrons in the metal wire is