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Electromagnetic Induction question

2024 · 1 Feb · Shift 1 · Q81
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Electromagnetic Induction question

2024 · 1 Feb · Shift 1 · Q81

JEE MainPhysicsElectromagnetic InductionNumerical+4 / −1
A rectangular loop of sides 12 cm12 \mathrm{~cm}12 cm and 5 cm5 \mathrm{~cm}5 cm, with its sides parallel to the xxx-axis and yyy-axis respectively, moves with a velocity of 5 cm/s5 \mathrm{~cm} / \mathrm{s}5 cm/s in the positive xxx axis direction, in a space containing a variable magnetic field in the positive zzz direction. The field has a gradient of 10−3 T/cm10^{-3} \mathrm{~T} / \mathrm{cm}10−3 T/cm along the negative xxx direction and it is decreasing with time at the rate of 10−3 T/s10^{-3} \mathrm{~T} / \mathrm{s}10−3 T/s. If the resistance of the loop is 6 mΩ6 \mathrm{~m} \Omega6 mΩ, the power dissipated by the loop as heat is ‾×10−9 W\underline{\hspace{2cm}}\times 10^{-9} \mathrm{~W}​×10−9 W.
Numerical answer
View written solutionFree

Correct answer: 216

  1. Given data
  • Rectangle dimensions: lx=12 cm=0.12 m,ly=5 cm=0.05 ml_x = 12\text{ cm} = 0.12\text{ m}, \qquad l_y = 5\text{ cm} = 0.05\text{ m}lx​=12 cm=0.12 m,ly​=5 cm=0.05 m
  • Area of loop: A=12×5=60 cm2=6×10−3 m2A = 12\times 5 = 60\text{ cm}^2 = 6\times 10^{-3}\text{ m}^2A=12×5=60 cm2=6×10−3 m2
  • Velocity of loop along +x+x+x: v=5 cm/s=0.05 m/sv = 5\text{ cm/s} = 0.05\text{ m/s}v=5 cm/s=0.05 m/s
  • Magnetic field is along +z+z+z.
  • Field gradient along negative xxx direction: dBdx=−10−3 T/cm\frac{dB}{dx} = -10^{-3}\,\text{T/cm}dxdB​=−10−3T/cm Since the field decreases as xxx increases, magnitude of gradient is ∣dBdx∣=10−3 T/cm=0.1 T/m\left|\frac{dB}{dx}\right| = 10^{-3}\,\text{T/cm} = 0.1\,\text{T/m}​dxdB​​=10−3T/cm=0.1T/m
  • Field is also decreasing with time: ∂B∂t=−10−3 T/s\frac{\partial B}{\partial t} = -10^{-3}\,\text{T/s}∂t∂B​=−10−3T/s
  • Resistance: R=6 mΩ=6×10−3 ΩR = 6\,\text{m}\Omega = 6\times 10^{-3}\,\OmegaR=6mΩ=6×10−3Ω
  1. Change of magnetic field experienced by the moving loop

The magnetic field changes for two reasons:

  • explicit time variation: ∂B∂t\dfrac{\partial B}{\partial t}∂t∂B​
  • motion of loop into region of different field: v∂B∂xv\dfrac{\partial B}{\partial x}v∂x∂B​

Hence total rate of change of field experienced by the loop is dBdt=∂B∂t+v∂B∂x\frac{dB}{dt} = \frac{\partial B}{\partial t} + v\frac{\partial B}{\partial x}dtdB​=∂t∂B​+v∂x∂B​

Now, v∂B∂x=(5 cm/s)(−10−3 T/cm)=−5×10−3 T/sv\frac{\partial B}{\partial x} = (5\,\text{cm/s})(-10^{-3}\,\text{T/cm}) = -5\times 10^{-3}\,\text{T/s}v∂x∂B​=(5cm/s)(−10−3T/cm)=−5×10−3T/s

Therefore, dBdt=−10−3−5×10−3=−6×10−3 T/s\frac{dB}{dt} = -10^{-3} - 5\times 10^{-3} = -6\times 10^{-3}\,\text{T/s}dtdB​=−10−3−5×10−3=−6×10−3T/s

So magnitude is ∣dBdt∣=6×10−3 T/s\left|\frac{dB}{dt}\right| = 6\times 10^{-3}\,\text{T/s}​dtdB​​=6×10−3T/s

  1. Induced emf

Magnetic flux through the loop is Φ=BA\Phi = BAΦ=BA

Thus induced emf magnitude is E=A∣dBdt∣\mathcal{E} = A\left|\frac{dB}{dt}\right|E=A​dtdB​​

So, E=(60 cm2)(6×10−3 T/s)\mathcal{E} = (60\,\text{cm}^2)(6\times 10^{-3}\,\text{T/s})E=(60cm2)(6×10−3T/s)

Using SI units: E=(6×10−3)(6×10−3)=36×10−6 V=3.6×10−5 V\mathcal{E} = (6\times 10^{-3})(6\times 10^{-3}) = 36\times 10^{-6}\,\text{V} = 3.6\times 10^{-5}\,\text{V}E=(6×10−3)(6×10−3)=36×10−6V=3.6×10−5V

  1. Current induced in the loop

I=ER=3.6×10−56×10−3=6×10−3 AI = \frac{\mathcal{E}}{R} = \frac{3.6\times 10^{-5}}{6\times 10^{-3}} = 6\times 10^{-3}\,\text{A}I=RE​=6×10−33.6×10−5​=6×10−3A

  1. Power dissipated as heat

P=I2R=E2RP = I^2R = \frac{\mathcal{E}^2}{R}P=I2R=RE2​

Using E2/R\mathcal{E}^2/RE2/R: P=(3.6×10−5)26×10−3P = \frac{(3.6\times 10^{-5})^2}{6\times 10^{-3}}P=6×10−3(3.6×10−5)2​

P=12.96×10−106×10−3P = \frac{12.96\times 10^{-10}}{6\times 10^{-3}}P=6×10−312.96×10−10​

P=2.16×10−7 WP = 2.16\times 10^{-7}\,\text{W}P=2.16×10−7W

Now write in the form asked: P=216×10−9 WP = 216\times 10^{-9}\,\text{W}P=216×10−9W

  1. Final answer

The required integer is 216\boxed{216}216​

This matches the stored correct answer.

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