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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2024
  • Hard
  • Numerical

A rectangular loop of sides \(12 \mathrm{~cm}\) and \(5 \mathrm{~cm}\), with its sides parallel to the \(x\)-axis and \(y\)-axis respectively moves with a velocity of \(5 \mathrm{~cm} / \mathrm{s}\) in the positive \(\mathrm{x}\) axis direction, in a space containing a variable magnetic field in the positive \(z\) direction. The field has a gradient of \(10^{-3} \mathrm{~T} / \mathrm{cm}\) along the negative \(\mathrm{x}\) direction and it is decreasing with time at the rate of \(10^{-5} \mathrm{~T} / \mathrm{s}\). If the resistance of the loop is \(6 \mathrm{~m} \Omega\), the power dissipated by the loop as heat is _______ \(\times 10^{-9} \mathrm{~W}\).

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Answer & Solution

Correct Answer

(B) \(216\)

Step-by-step Solution

Detailed explanation

\(\mathrm{B}_0\) is the magnetic field at origin

\(\frac{d B}{d x}=-\frac{10^{-3}}{10^{-2}}\)

\(\int_{B_0}^B d B=-\int_0^x 10^{-1} d x\)

\(B-B_0=-10^{-1} x\)

\(B=\left(B_0-\frac{x}{10}\right)\)

Motional emf in \(\mathrm{AB}=0\)

Motional emf in \(\mathrm{CD}=0\)

Motional emf in \(\mathrm{AD}=\varepsilon_1=\mathrm{B}_0 / \mathrm{v}\)

Magnetic field on \(\operatorname{rod} B C B\)

\(=\left(\mathrm{B}_0-\frac{\left(-12 \times 10^{-2}\right)}{10}\right)\)

Motional emf in \(\mathrm{BC}=\varepsilon_2=\left(\mathrm{B}_0+\frac{12 \times 10^{-2}}{10}\right) \ell \times \mathrm{v}\)

\(\varepsilon_{\text {eq }}=\varepsilon_2-\varepsilon_1=300 \times 10^{-7} \mathrm{~V}\)

For time variation

\(\left(\varepsilon_{\text {eq }}\right)^{\prime}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=60 \times 10^{-7} \mathrm{~V}\)

\(\left(\varepsilon_{\text {eq }}\right)_{\text {net }}=\varepsilon_{\text {eq }}+\left(\varepsilon_{\text {eq }}\right)^{\prime}=360 \times 10^{-7} \mathrm{~V}\)

\(\text { Power }=\frac{\left(\varepsilon_{\text {eq }}\right)_{\text {net }}^2}{\mathrm{R}}=216 \times 10^{-9} \mathrm{~W}\)

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