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Electromagnetic Induction question

2023 · 29 Jan · Shift 2 · Q48
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  5. /2023 · 29 Jan · Shift 2 · Q48

Electromagnetic Induction question

2023 · 29 Jan · Shift 2 · Q48

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
A square loop of area 25 cm 2^22 has a resistance of 10 Ω\OmegaΩ. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be
  1. A
    1.0×10−3 J\mathrm{1.0\times10^{-3}~J}1.0×10−3 J
  2. B
    5×10−3 J\mathrm{5\times10^{-3}~J}5×10−3 J
  3. C
    2.5×10−3 J\mathrm{2.5\times10^{-3}~J}2.5×10−3 J
  4. D
    1.0×10−4 J\mathrm{1.0\times10^{-4}~J}1.0×10−4 J
View written solutionFree

Correct answer: A

  1. Given data
  • Area of square loop: A=25 cm2=25×10−4 m2=2.5×10−3 m2A=25~\text{cm}^2=25\times 10^{-4}~\text{m}^2=2.5\times 10^{-3}~\text{m}^2A=25 cm2=25×10−4 m2=2.5×10−3 m2
  • Resistance of loop: R=10 ΩR=10~\OmegaR=10 Ω
  • Magnetic field: B=40.0 TB=40.0~\text{T}B=40.0 T
  • Time taken to pull loop completely out of field: t=1.0 st=1.0~\text{s}t=1.0 s

The plane of the loop is perpendicular to the magnetic field, so initially the magnetic flux is maximum.

  1. Initial and final magnetic flux

Initially, the loop is completely inside the field, so Φi=BA\Phi_i=BAΦi​=BA

Finally, the loop is completely outside the field, so Φf=0\Phi_f=0Φf​=0

Hence change in flux is ΔΦ=BA\Delta \Phi=BAΔΦ=BA

So the induced emf during uniform withdrawal is E=ΔΦΔt=BAt\mathcal{E}=\frac{\Delta \Phi}{\Delta t}=\frac{BA}{t}E=ΔtΔΦ​=tBA​

Substitute values: E=40×2.5×10−31=0.1 V\mathcal{E}=\frac{40\times 2.5\times 10^{-3}}{1}=0.1~\text{V}E=140×2.5×10−3​=0.1 V

  1. Induced current

Using Ohm’s law, I=ER=0.110=0.01 AI=\frac{\mathcal{E}}{R}=\frac{0.1}{10}=0.01~\text{A}I=RE​=100.1​=0.01 A

  1. Work done in pulling the loop out

Since the loop is pulled slowly and uniformly, the external work done is equal to the electrical energy dissipated as heat in the resistance: W=I2RtW=I^2RtW=I2Rt

Substitute values: W=(0.01)2×10×1W=(0.01)^2\times 10\times 1W=(0.01)2×10×1 W=10−4×10=10−3 JW=10^{-4}\times 10=10^{-3}~\text{J}W=10−4×10=10−3 J

Therefore, W=1.0×10−3 J\boxed{W=1.0\times 10^{-3}~\text{J}}W=1.0×10−3 J​

  1. Option check
  • A: 1.0×10−3 J1.0\times 10^{-3}~\text{J}1.0×10−3 J ✅
  • B: 5×10−3 J5\times 10^{-3}~\text{J}5×10−3 J ❌
  • C: 2.5×10−3 J2.5\times 10^{-3}~\text{J}2.5×10−3 J ❌
  • D: 1.0×10−4 J1.0\times 10^{-4}~\text{J}1.0×10−4 J ❌

So the correct option is A.

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