A closed coil of 50 turns and area 0.02 m2 has a resistance of 5 Ω. Its plane is normal to a uniform magnetic field. The field rises steadily from zero to 0.5 T in 0.1 s. The heat produced in the coil during this time is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(0.5\ \mathrm{J}\)
The field rises at a steady rate, so the induced emf is constant. The induced current \( I=\varepsilon/R \) heats the coil. In time \( t \) the heat is \( I^{2}Rt \), which is the same as \( \varepsilon^{2}t/R \).
\( N=50 \)
\( A=0.02\ \mathrm{m^{2}} \)
\( \Delta B=0.5\ \mathrm{T} \)
\( t=0.1\ \mathrm{s} \)
\( R=5\ \Omega \)
\( \begin{aligned} \varepsilon &= \frac{NA\,\Delta B}{t} \\ &= \frac{50\times0.02\times0.5}{0.1} \\ &= 5\ \mathrm{V} \end{aligned} \)
\( \begin{aligned} H &= \frac{\varepsilon^{2}t}{R} \\ &= \frac{5^{2}\times0.1}{5} \\ &= 0.5\ \mathrm{J} \end{aligned} \)
NCERT (2026-27):
Faraday’s Law of Induction – “In the case of a closely wound coil of N turns, change of flux associated with each turn, is the same. Therefore, the expression for the total induced emf is given by”
\( \displaystyle \varepsilon=-N\frac{d\Phi_B}{dt} \)\( \qquad (6.4) \)
Summary – “… If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.”
Lenz’s Law and Conservation of Energy – “… This energy is dissipated by Joule heating produced by the induced current.”