Induced EMF in a rotating loop is given by
$$ e = B A \omega \sin\theta $$
Where $A=\pi r^2$
Given:
$$ e = 15.4\times10^{-3}\ \text{V},\quad B = 0.5\ \text{T},\quad \omega = 100\ \text{rad s}^{-1},\quad \theta = 30^\circ $$
Substituting:
$$ 15.4\times10^{-3} = 0.5 \times \pi r^2 \times 100 \times \sin30^\circ $$
$$ 15.4\times10^{-3} = 25\pi r^2 $$
$$ r^2 = \frac{15.4\times10^{-3}}{25\pi} $$
Using $\pi=\dfrac{22}{7}$,
$$ r = 0.014\ \text{m} = 14\ \text{mm} $$
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.