JEE Main 2023PhysicsCurrent ElectricityEasyNumerical

JEE Main 2023 — Current Electricity Question with Solution

JEE Main 2023 (08 Apr Shift 1)

Question

A current of 2 A flows through a wire of cross-sectional area 25.0 mm2. The number of free electrons in a cubic meter are 2.0×1028. The drift velocity of the electrons is _____ ×10-6 ms-1

(given, charge on electron =1.6×10-19C).

Enter your answer

Show full solutionCorrect answer: 25
Correct answer
25

Step-by-step explanation

The drift velocity is given by 

vd=IneA

⇒I=nAevd   ...(i)

The given data is 

I=2 AA=25×10-6 m2n=2×1028e=1.6×10-19 C

Substituting the given values in equation (i)

2=2×1028×25×10−6×1.6×10−19×vd⇒vd=22×1028×25×10−6×1.6×10−19

⇒vd=125×1.6×1000 m s-1=25×10−6 m s-1

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About this question

This is a previous-year question from JEE Main 2023, covering the Current Electricity chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.