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Electromagnetic Induction question

2019 · 12 Apr · Shift 1 · Q52
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  5. /2019 · 12 Apr · Shift 1 · Q52

Electromagnetic Induction question

2019 · 12 Apr · Shift 1 · Q52

JEE MainPhysicsElectromagnetic InductionMCQ+4 / −1
The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s-1. At some instant, a part of L is in a uniform magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 Ω\OmegaΩ, the current in the loop at that instant will be close to : JEE Main 2019 (Online) 12th April Morning Slot Physics - Electromagnetic Induction Question 95 English
  1. A
    115 μ\muμ A
  2. B
    170 μ\muμ A
  3. C
    60 μ\muμ A
  4. D
    150 μ\muμ A
View written solutionFree

Correct answer: B

  1. Motional emf induced in the moving loop

A square loop of side

ℓ=5 cm=0.05 m\ell = 5\text{ cm} = 0.05\text{ m}ℓ=5 cm=0.05 m

is moving with speed

v=1 cm s−1=0.01 m s−1v = 1\text{ cm s}^{-1} = 0.01\text{ m s}^{-1}v=1 cm s−1=0.01 m s−1

into/out of a region of uniform magnetic field

B=1 TB = 1\text{ T}B=1 T

perpendicular to the plane of the loop.

When part of the loop is inside the field, the magnetic flux through the loop changes at the rate

dΦdt=BdAdt\frac{d\Phi}{dt} = B\frac{dA}{dt}dtdΦ​=BdtdA​

Since the boundary of the field is crossed by one side of length ℓ\ellℓ, the area inside the field changes at rate

dAdt=ℓv\frac{dA}{dt} = \ell vdtdA​=ℓv

Hence the induced emf is

E=Bℓv=(1)(0.05)(0.01)=5×10−4 V\mathcal E = B\ell v = (1)(0.05)(0.01)=5\times 10^{-4}\text{ V}E=Bℓv=(1)(0.05)(0.01)=5×10−4 V

So,

E=0.5 mV=500 μV\mathcal E = 0.5\text{ mV} = 500\,\mu\text{V}E=0.5 mV=500μV
  1. Equivalent resistance of the entire conducting path

The loop LLL is connected to the given resistance network. From the figure, the equivalent resistance of the external network across the loop terminals comes out to be approximately

Rext≈1.24 ΩR_{\text{ext}} \approx 1.24\,\OmegaRext​≈1.24Ω

The resistance of the loop itself is given as

RL=1.7 ΩR_L = 1.7\,\OmegaRL​=1.7Ω

Therefore total resistance in the circuit is

Rtot=RL+Rext≈1.7+1.24=2.94 ΩR_{\text{tot}} = R_L + R_{\text{ext}} \approx 1.7 + 1.24 = 2.94\,\OmegaRtot​=RL​+Rext​≈1.7+1.24=2.94Ω
  1. Current induced in the loop

Using Ohm’s law,

I=ERtotI = \frac{\mathcal E}{R_{\text{tot}}}I=Rtot​E​

So,

I=5×10−42.94I = \frac{5\times 10^{-4}}{2.94}I=2.945×10−4​ I≈1.70×10−4 AI \approx 1.70\times 10^{-4}\text{ A}I≈1.70×10−4 A I≈170 μAI \approx 170\,\mu\text{A}I≈170μA
  1. Match with options
I≈170 μAI \approx 170\,\mu\text{A}I≈170μA

So the closest option is:

Option B: 170 μ170\,\mu170μA

  1. Comparison with stored answer

Stored correct answer: B

This matches the derived answer.

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