Two circular loops are made from the same wire. Their radii are \(r\) and \(2r\). Both loops are placed with their planes normal to the same uniform magnetic field, which changes at a steady rate. The ratio of the induced current in the smaller loop to that in the larger loop is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(1:2\)
The emf in a loop depends on its area, and the current also depends on its resistance. Both loops are made from the same wire, so the resistance of a loop is proportional to its length \( 2\pi r \). Let \( \lambda \) be the resistance per unit length of the wire.
\( \begin{aligned} \varepsilon &= \pi r^{2}\frac{dB}{dt} \\ R &= 2\pi r\lambda \end{aligned} \)
The induced current is
\( \begin{aligned} I &= \frac{\varepsilon}{R} \\ &= \frac{\pi r^{2}(dB/dt)}{2\pi r\lambda} \\ &= \frac{r}{2\lambda}\frac{dB}{dt} \end{aligned} \)
So \( I \) is proportional to \( r \).
\( \begin{aligned} \frac{I_1}{I_2} &= \frac{r}{2r} \\ &= \frac{1}{2} \end{aligned} \)
Hence, \( I_1:I_2=1:2 \).
NCERT (2026-27):
Faraday’s Law of Induction – “Mathematically, the induced emf is given by”
\( \displaystyle \varepsilon=-\frac{d\Phi_B}{dt} \)\( \qquad (6.3) \)
Summary – “… If the circuit is closed, a current I = ε/R is set up in it, where R is the resistance of the circuit.”