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QJEEM24Ph8e59: JEE Main 2024 Physics - Class 12 JEEM24Ph8e59 Solution: A square loop PQRS having 10 turns, area and resistance is…

A square loop PQRS having 10 turns, area 3.6×10−3 m23.6 \times 10^{-3} \mathrm{~m}^2 and resistance 100Ω100 \Omega is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B=0.5 T\mathrm{B}=0.5 \mathrm{~T} as shown. Work done in pulling the loop out of the field in 1.0 s1.0 \mathrm{~s} is _________ ×10−6 J\times 10^{-6} \mathrm{~J} .

JEE Main 2024 (Online) 8th April Morning Shift Physics - Magnetic Effect of Current Question 20 English

Step-by-Step Verified Solution

amp;A=36×10−4 m2amp;I=6×10−2 mamp;=6 cmVamp;=6 cm1sec=6 cm/sεamp;=B/vn2=0.5×6100×6100amp;=18×10−4 VEamp;=n2ε2Rt=100×18×18×10−4×10−4102×1amp;=324×10−10×102amp;=3.24×10−6\begin{aligned}& A = 36 \times 10^{-4} \mathrm{~m}^2 \\& I= 6 \times 10^{-2} \mathrm{~m} \\& =6 \mathrm{~cm} \\V & =\frac{6 \mathrm{~cm}}{1 \mathrm{sec}}=6 \mathrm{~cm} / \mathrm{s} \\\varepsilon & =B / \mathrm{vn}^2=0.5 \times \frac{6}{100} \times \frac{6}{100} \\& =18 \times 10^{-4} \mathrm{~V} \\E & =\frac{n^2 \varepsilon^2}{R} t=100 \times \frac{18 \times 18 \times 10^{-4} \times 10^{-4}}{10^2} \times 1 \\& =324 \times 10^{-10} \times 10^2 \\& =3.24 \times 10^{-6}\end{aligned}

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