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Physics · Electromagnetic Induction

JEE Main 2024 — 1 February, Shift 1 — Question 48

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

Answer: 216

Numerical answer — enter this value.

Step-by-step solution

B0B_{0} is the magnetic field at origin dBdx=−10−310−2\frac{\mathrm{dB}}{\mathrm{dx}}=-\frac{10^{-3}}{10^{-2}}

∫B0BdB=−∫0x10−1dx\int_{B_{0}}^{B} d B=-\int_{0}^{x} 10^{-1} d x B−B0=−10−1x\mathrm{B}-\mathrm{B}_{0}=-10^{-1} \mathrm{x}

B=(B0−x10)B=\left(B_{0}-\frac{\mathrm{x}}{10}\right) Motional emf in AB=0\mathrm{AB}=0 Motional emf in CD=0\mathrm{CD}=0

Motional emf in AD=ε1=B0ℓv\mathrm{AD}=\varepsilon_{1}=\mathrm{B}_{0} \ell \mathrm{v} Magnetic field on rod BC B =(B0−(−12×10−2)10)=\left(\mathrm{B}_{0}-\frac{\left(-12 \times 10^{-2}\right)}{10}\right)

Motional emf in BC=ε2=(B0+12×10−210)ℓ×v\mathrm{BC}=\varepsilon_{2}=\left(\mathrm{B}_{0}+\frac{12 \times 10^{-2}}{10}\right) \ell \times \mathrm{v} εeq=ε2−ε1=300×10−7 V\varepsilon_{\mathrm{eq}}=\varepsilon_{2}-\varepsilon_{1}=300 \times 10^{-7} \mathrm{~V}

For time variation

(εeq)′=AdBdt=60×10−7 V\left(\varepsilon_{\mathrm{eq}}\right)^{\prime}=\mathrm{A} \frac{\mathrm{dB}}{\mathrm{dt}}=60 \times 10^{-7} \mathrm{~V} (εeq )net =εeq +(εeq )′=360×10−7 V\left(\varepsilon_{\text {eq }}\right)_{\text {net }}=\varepsilon_{\text {eq }}+\left(\varepsilon_{\text {eq }}\right)^{\prime}=360 \times 10^{-7} \mathrm{~V}

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

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
1 February, Shift 1
Subject
Physics
Chapter
Electromagnetic Induction
Topic
Motional EMF