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Magnetics question

2024 · 8 Apr · Shift 1 · Q84
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Magnetics question

2024 · 8 Apr · Shift 1 · Q84

JEE MainPhysicsMagneticsNumerical+4 / −1
A square loop PQRS having 10 turns, area 3.6×10−3 m23.6 \times 10^{-3} \mathrm{~m}^23.6×10−3 m2 and resistance 100Ω100 \Omega100Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B=0.5 T\mathrm{B}=0.5 \mathrm{~T}B=0.5 T as shown. Work done in pulling the loop out of the field in 1.0 s1.0 \mathrm{~s}1.0 s is ‾\underline{\hspace{2cm}}​×10−6 J\times 10^{-6} \mathrm{~J}×10−6 J. JEE Main 2024 (Online) 8th April Morning Shift Physics - Magnetic Effect of Current Question 30 English
Numerical answer
View written solutionFree

Correct answer: 3

  1. Given data
  • Number of turns: N=10N=10N=10
  • Area of square loop: A=3.6×10−3 m2A=3.6\times 10^{-3}\,\text{m}^2A=3.6×10−3m2
  • Resistance: R=100 ΩR=100\,\OmegaR=100Ω
  • Magnetic field: B=0.5 TB=0.5\,\text{T}B=0.5T
  • Time taken to pull it out: t=1.0 st=1.0\,\text{s}t=1.0s

Since the loop is slowly and uniformly pulled completely out of the field, the magnetic flux linked with the loop decreases uniformly from its initial value to zero.

  1. Initial flux linkage

For one turn, initial flux is

ϕ=BA\phi = BAϕ=BA

So for NNN turns, initial flux linkage is

Nϕ=NBAN\phi = NBANϕ=NBA

Substituting values:

NBA=10×0.5×3.6×10−3NBA = 10\times 0.5 \times 3.6\times 10^{-3}NBA=10×0.5×3.6×10−3 =18×10−3= 18\times 10^{-3}=18×10−3 =1.8×10−2 Wb-turn= 1.8\times 10^{-2}\,\text{Wb-turn}=1.8×10−2Wb-turn
  1. Induced emf

Average induced emf:

E=Δ(Nϕ)Δt=NBAt\mathcal{E} = \frac{\Delta(N\phi)}{\Delta t} = \frac{NBA}{t}E=ΔtΔ(Nϕ)​=tNBA​

Thus,

E=1.8×10−21=1.8×10−2 V\mathcal{E} = \frac{1.8\times 10^{-2}}{1} = 1.8\times 10^{-2}\,\text{V}E=11.8×10−2​=1.8×10−2V
  1. Induced current

Using Ohm’s law,

I=ER=1.8×10−2100I = \frac{\mathcal{E}}{R} = \frac{1.8\times 10^{-2}}{100}I=RE​=1001.8×10−2​ =1.8×10−4 A= 1.8\times 10^{-4}\,\text{A}=1.8×10−4A
  1. Work done in pulling the loop out

Since the motion is slow and uniform, the mechanical work done equals the electrical energy dissipated as heat:

W=I2RtW = I^2RtW=I2Rt

Substitute values:

W=(1.8×10−4)2×100×1W = (1.8\times 10^{-4})^2 \times 100 \times 1W=(1.8×10−4)2×100×1 =3.24×10−8×100= 3.24\times 10^{-8} \times 100=3.24×10−8×100 =3.24×10−6 J= 3.24\times 10^{-6}\,\text{J}=3.24×10−6J
  1. Final answer
W=3.24×10−6 JW = 3.24\times 10^{-6}\,\text{J}W=3.24×10−6J

So the required integer in the blank is:

3\boxed{3}3​
  1. Comparison with stored answer

Stored correct answer = 333

Our derived answer = 333

Hence, the derived answer agrees with the stored answer.

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