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Current Electricity question

2023 · 31 Jan · Shift 1 · Q53
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  5. /2023 · 31 Jan · Shift 1 · Q53

Current Electricity question

2023 · 31 Jan · Shift 1 · Q53

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
The drift velocity of electrons for a conductor connected in an electrical circuit is Vd\mathrm{V}_{\mathrm{d}}Vd​. The conductor in now replaced by another conductor with same material and same length but double the area of cross section. The applied voltage remains same. The new drift velocity of electrons will be
  1. A
    Vd4\frac{V_{d}}{4}4Vd​​
  2. B
    Vd\mathrm{V}_{\mathrm{d}}Vd​
  3. C
    2 Vd2 \mathrm{~V}_{\mathrm{d}}2 Vd​
  4. D
    Vd2\frac{V_{d}}{2}2Vd​​
View written solutionFree

Correct answer: B

  1. Use the relation for drift velocity

For a conductor, I=nAevdI = n A e v_dI=nAevd​ where:

  • III = current
  • nnn = number density of free electrons
  • AAA = cross-sectional area
  • eee = charge of electron
  • vdv_dvd​ = drift velocity

Also, by Ohm’s law, I=VRI = \frac{V}{R}I=RV​ and resistance is R=ρLAR = \rho \frac{L}{A}R=ρAL​

So, I=VρL/A=VAρLI = \frac{V}{\rho L/A} = \frac{VA}{\rho L}I=ρL/AV​=ρLVA​

  1. Find drift velocity in terms of applied voltage

Substitute III into I=nAevdI = nAe v_dI=nAevd​: nAevd=VAρLnAe v_d = \frac{VA}{\rho L}nAevd​=ρLVA​

Cancel AAA from both sides: nevd=VρLne v_d = \frac{V}{\rho L}nevd​=ρLV​

Thus, vd=VneρLv_d = \frac{V}{n e \rho L}vd​=neρLV​

  1. Analyze the changed conductor

The new conductor has:

  • same material ⇒ρ\Rightarrow \rho⇒ρ same, nnn same
  • same length LLL
  • double area A′=2AA' = 2AA′=2A
  • same applied voltage VVV

From vd=VneρLv_d = \frac{V}{n e \rho L}vd​=neρLV​ we see that vdv_dvd​ does not depend on area.

Hence the new drift velocity remains unchanged: vd′=vdv_d' = v_dvd′​=vd​

  1. Check options
  • A: Vd4\dfrac{V_d}{4}4Vd​​ — incorrect
  • B: VdV_dVd​ — correct
  • C: 2Vd2V_d2Vd​ — incorrect
  • D: Vd2\dfrac{V_d}{2}2Vd​​ — incorrect

Therefore, the correct answer is B.

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