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JEE Main 2019 (12 Apr, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2019 (12 Apr, Shift 1)

Q48JEE Main 2019 · 12 APR · SHIFT 1Difficult

Motional EMF

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\Omega, the current in the loop at that instant will be close to:

Figure

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Answer: (b)

  1. emf in loop: e=Blv=1×0.05×0.01=5×10−4 Ve = Blv = 1 \times 0.05 \times 0.01 = 5 \times 10^{-4}\text{ V}
  2. Bridge fed at B and D is symmetric top-bottom, so A and C are at the same potential: no current in the 3 Ω3\ \Omega.
  3. RBD=(1+1)∥(2+2)=2×46=43 ΩR_{BD} = (1+1) \parallel (2+2) = \dfrac{2 \times 4}{6} = \dfrac{4}{3}\ \Omega
  4. Total R=1.7+1.33=3.03 ΩR = 1.7 + 1.33 = 3.03\ \Omega, I=5×10−43.03≈1.65×10−4 A≈170 μAI = \dfrac{5 \times 10^{-4}}{3.03} \approx 1.65 \times 10^{-4}\text{ A} \approx 170\ \mu\text{A}
  5. So the answer is (b) 170 μ\muA.
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