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JEE Main 2024 (1 Feb, Shift 1) · Motional EMF

Motional EMF question from JEE Main 2024 (1 Feb, Shift 1)

Q22JEE Main 2024 · 1 FEB · SHIFT 1Difficult

Motional EMF

A rectangular loop of sides 12 cm12\text{ cm} and 5 cm5\text{ cm}, with its sides parallel to the xx-axis and yy-axis respectively, moves with a velocity of 5 cm/s5\text{ cm/s} in the positive xx axis direction, in a space containing a variable magnetic field in the positive zz direction. The field has a gradient of 10−3 T/cm10^{-3}\text{ T/cm} along the negative xx direction and it is decreasing with time at the rate of 10−3 T/s10^{-3}\text{ T/s}. If the resistance of the loop is 6 mΩ6\text{ m}\Omega, the power dissipated by the loop as heat is __________ ×10−9 W\times 10^{-9}\text{ W}.

Numerical answer type. Work it out, then open the answer.

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Answer: 216

  1. Rate of change of BB seen by the moving loop: dBdt=∂B∂t+v∂B∂x\dfrac{dB}{dt} = \dfrac{\partial B}{\partial t} + v\dfrac{\partial B}{\partial x}
  2. ∂B∂x=−10−3 T/cm=−0.1 T/m\dfrac{\partial B}{\partial x} = -10^{-3}\text{ T/cm} = -0.1\text{ T/m}, so dBdt=−10−3+0.05×(−0.1)=−6×10−3 T/s\dfrac{dB}{dt} = -10^{-3} + 0.05 \times (-0.1) = -6 \times 10^{-3}\text{ T/s}
  3. Area A=0.12×0.05=6×10−3 m2A = 0.12 \times 0.05 = 6 \times 10^{-3}\text{ m}^2, so e=A∣dBdt∣=3.6×10−5 Ve = A\left|\dfrac{dB}{dt}\right| = 3.6 \times 10^{-5}\text{ V}
  4. P=e2R=1.296×10−96×10−3=2.16×10−7 WP = \dfrac{e^2}{R} = \dfrac{1.296 \times 10^{-9}}{6 \times 10^{-3}} = 2.16 \times 10^{-7}\text{ W}
  5. So P=216×10−9 WP = 216 \times 10^{-9}\text{ W}, answer 216.
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