Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Current Electricity question

2019 · 10 Apr · Shift 1 · Q61
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Current Electricity
  5. /2019 · 10 Apr · Shift 1 · Q61

Current Electricity question

2019 · 10 Apr · Shift 1 · Q61

JEE MainPhysicsCurrent ElectricityMCQ+4 / −1
A current of 5 A passes through a copper conductor (resistivity = 1.7 × 10–8 Ω\OmegaΩ m) of radius of cross-section 5 mm. Find the mobility of the charges if their drift velocity is 1.1 × 10–3 m/s.
  1. A
    1.3 m2/Vs
  2. B
    1.0 m2/Vs
  3. C
    1.8 m2/Vs
  4. D
    1.5 m2/Vs
View written solutionFree

Correct answer: B

  1. Given data
  • Current: I=5 AI = 5\,\text{A}I=5A
  • Resistivity of copper: ρ=1.7×10−8 Ω m\rho = 1.7 \times 10^{-8}\,\Omega\,\text{m}ρ=1.7×10−8Ωm
  • Radius of conductor: r=5 mm=5×10−3 mr = 5\,\text{mm} = 5 \times 10^{-3}\,\text{m}r=5mm=5×10−3m
  • Drift velocity: vd=1.1×10−3 m/sv_d = 1.1 \times 10^{-3}\,\text{m/s}vd​=1.1×10−3m/s

We need to find the mobility μ\muμ of charge carriers.

  1. Use relation between drift velocity and mobility

The drift velocity is related to electric field by

vd=μEv_d = \mu Evd​=μE

So,

μ=vdE\mu = \frac{v_d}{E}μ=Evd​​

Thus, we first need the electric field EEE inside the conductor.

  1. Find current density

Cross-sectional area:

A=πr2=π(5×10−3)2A = \pi r^2 = \pi (5 \times 10^{-3})^2A=πr2=π(5×10−3)2 A=π×25×10−6=25π×10−6 m2A = \pi \times 25 \times 10^{-6} = 25\pi \times 10^{-6}\,\text{m}^2A=π×25×10−6=25π×10−6m2

Current density:

J=IA=525π×10−6J = \frac{I}{A} = \frac{5}{25\pi \times 10^{-6}}J=AI​=25π×10−65​ J=15π×106≈6.37×104 A/m2J = \frac{1}{5\pi} \times 10^6 \approx 6.37 \times 10^4\,\text{A/m}^2J=5π1​×106≈6.37×104A/m2
  1. Use microscopic form of Ohm’s law

We know,

E=ρJE = \rho JE=ρJ

So,

E=(1.7×10−8)(6.37×104)E = (1.7 \times 10^{-8})(6.37 \times 10^4)E=(1.7×10−8)(6.37×104) E≈1.083×10−3 V/mE \approx 1.083 \times 10^{-3}\,\text{V/m}E≈1.083×10−3V/m
  1. Calculate mobility
μ=vdE=1.1×10−31.083×10−3\mu = \frac{v_d}{E} = \frac{1.1 \times 10^{-3}}{1.083 \times 10^{-3}}μ=Evd​​=1.083×10−31.1×10−3​ μ≈1.02 m2/V s\mu \approx 1.02\,\text{m}^2/\text{V s}μ≈1.02m2/V s

So the mobility is approximately

1.0 m2/V s\boxed{1.0\,\text{m}^2/\text{V s}}1.0m2/V s​
  1. Check options
  • A: 1.3 m2/V s1.3\,\text{m}^2/\text{V s}1.3m2/V s
  • B: 1.0 m2/V s1.0\,\text{m}^2/\text{V s}1.0m2/V s
  • C: 1.8 m2/V s1.8\,\text{m}^2/\text{V s}1.8m2/V s
  • D: 1.5 m2/V s1.5\,\text{m}^2/\text{V s}1.5m2/V s

Hence, the correct option is:

B\boxed{\text{B}}B​
PreviousNext

More from Current Electricity

  • In the given circuit, an ideal voltmeter connected across the 10 Ω resistance reads 2V. The internal resistance r, of each cell is: Includes diagram2019 · MCQ
  • Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity ρ. The resistance between the two spheres will be :2019 · MCQ
  • A 2 W carbon resistor is color coded with green, black, red and brown respectively. The maximum current which can be passed through this resistor is -2019 · MCQ
  • A uniform metallic wire has a resistance of 18 Ω and is bent into an equilateral triangle. Then, the resistance between any two vertices of the triangle is -2019 · MCQ
  • In the given circuit the cells have zero internal resistance. The currents (in Amperes) passing through resistance R1 and R2 respectively, are - Includes diagram2019 · MCQ
  • The actual value of resistance R, shown in the figure is 30 Ω. This is measuered in an experiment as shown using the standard formula R = IV​, where V and I are the readings of the voltmeter and ammeter,… Includes diagram2019 · MCQ
  • The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80 Ω and 40 Ω, respectively. Assuming that the colour code… Includes diagram2019 · MCQ
  • A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is -2019 · MCQ