A square loop PQRS having 10 turns, area 3.6 × 10⁻³ m² and resistance 100 Ω is slowly and uniformly being pulled out of a uniform magnetic field of magnitude B = 0.5 T as shown. Work done in pulling the loop out of the field in 1.0 s is _________ × 10⁻⁶ J. Numerical answer type. Work it out, then open the answer.

Answer: 3. Side L = √3.6 × 10⁻³ = 0.06 m; pulled out in 1 s, so v = 0.06 m/s. emf e = NBLv = 10 × 0.5 × 0.06 × 0.06 = 0.018 V Work = heat produced = e²/R t = ((0.018)²)/(100) × 1 = 3.24 × 10⁻⁶ J So the answer is about 3 × 10⁻⁶ J, i.e. 3.

A square loop PQRS having 10 turns, area 3.6 × 10⁻³ m² and resistance… | JEE Main 2024 (8 Apr, Shift 1)

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Other: Induction & Displacement Current

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

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

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Answer: 3

Side $L = \sqrt{3.6 \times 10^{-3}} = 0.06$ m; pulled out in $1$ s, so $v = 0.06$ m/s.
emf $e = NBLv = 10 \times 0.5 \times 0.06 \times 0.06 = 0.018$ V
Work = heat produced $= \frac{e^2}{R} t = \frac{(0.018)^2}{100} \times 1 = 3.24 \times 10^{-6}$ J
So the answer is about $3 \times 10^{-6}$ J, i.e. $3$ .

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