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Current Electricity question

2022 · 27 Jul · Shift 1 · Q67
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Current Electricity question

2022 · 27 Jul · Shift 1 · Q67

JEE MainPhysicsCurrent ElectricityNumerical+4 / −1
A 1 m1 \mathrm{~m}1 m long copper wire carries a current of 1 A1 \mathrm{~A}1 A. If the cross section of the wire is 2.0 mm22.0 \mathrm{~mm}^{2}2.0 mm2 and the resistivity of copper is 1.7×10−8 Ωm1.7 \times 10^{-8}\, \Omega \mathrm{m}1.7×10−8Ωm, the force experienced by moving electron in the wire is ‾×10−23 N\underline{\hspace{2cm}}\times 10^{-23} \mathrm{~N}​×10−23 N. (charge on electorn =1.6×10−19 C=1.6 \times 10^{-19} \,\mathrm{C}=1.6×10−19C)
Numerical answer
View written solutionFree

Correct answer: 136

  1. Find the electric field inside the wire

For a wire carrying steady current, V=IRV = IRV=IR and E=VL=IRLE = \frac{V}{L} = \frac{IR}{L}E=LV​=LIR​

Also, resistance of the wire is R=ρLAR = \frac{\rho L}{A}R=AρL​

So, E=IL⋅ρLA=ρIAE = \frac{I}{L}\cdot \frac{\rho L}{A} = \frac{\rho I}{A}E=LI​⋅AρL​=AρI​

  1. Substitute the given values

Given:

  • ρ=1.7×10−8 Ω m\rho = 1.7\times 10^{-8}\ \Omega\,\text{m}ρ=1.7×10−8 Ωm
  • I=1 AI = 1\ \text{A}I=1 A
  • A=2.0 mm2=2.0×10−6 m2A = 2.0\ \text{mm}^2 = 2.0\times 10^{-6}\ \text{m}^2A=2.0 mm2=2.0×10−6 m2

Hence, E=1.7×10−8×12.0×10−6E = \frac{1.7\times 10^{-8}\times 1}{2.0\times 10^{-6}}E=2.0×10−61.7×10−8×1​ E=0.85×10−2E = 0.85\times 10^{-2}E=0.85×10−2 E=8.5×10−3 N/CE = 8.5\times 10^{-3}\ \text{N/C}E=8.5×10−3 N/C

  1. Force on one electron

The magnitude of force on a moving electron due to electric field is F=eEF = eEF=eE

Given, e=1.6×10−19 Ce = 1.6\times 10^{-19}\ \text{C}e=1.6×10−19 C

So, F=1.6×10−19×8.5×10−3F = 1.6\times 10^{-19}\times 8.5\times 10^{-3}F=1.6×10−19×8.5×10−3 F=13.6×10−22F = 13.6\times 10^{-22}F=13.6×10−22 F=1.36×10−21 NF = 1.36\times 10^{-21}\ \text{N}F=1.36×10−21 N

  1. Write in the asked form

We need F=‾×10−23 NF = \underline{\hspace{0.5cm}}\times 10^{-23}\ \text{N}F=​×10−23 N

Now, 1.36×10−21=136×10−231.36\times 10^{-21} = 136\times 10^{-23}1.36×10−21=136×10−23

So the required integer is: 136\boxed{136}136​

  1. Comparison with stored answer

Stored correct answer = 136136136

Our derived answer also is 136136136, so it agrees.

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