JEE Main 2026 Jan 22 Shift 2, Physics Q21: Motional EMF

JEE Main2026Jan 22, Shift 2Physics

A conducting circular loop is rotated about its diameter at a constant angular speed of 100 rad/s in a magnetic field of 0.5 T perpendicular to the axis of rotation. When the loop is rotated by 30° from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is _____ mm. (Take $π = \frac{22}{7}$ )

Your answer

Solution

For a loop rotating in a magnetic field, $E M F = B A ω sin ω t$ . Here $ω t = 30°$ , so $sin ω t = \frac{1}{2}$ .

With $A = π r^{2}$ :

$15.4 \times 1 0^{- 3} = 0.5 \times \frac{22}{7} \times r^{2} \times 100 \times \frac{1}{2}$ .

$r^{2} = \frac{15.4 \times 1 0 ^{- 3} \times 7 \times 2}{0.5 \times 22 \times 100} = \frac{15.4 \times 14 \times 1 0 ^{- 3}}{1100} = \frac{15.4 \times 14 \times 1 0 ^{- 5}}{11}$ .

$r^{2} = 1.96 \times 1 0^{- 4}$ m $^{2}$ , so $r = 1.4 \times 1 0^{- 2}$ m $= 14$ mm.

Concept behind this question

Motional EMFNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Jan 22 Shift 2, Physics Q21: Motional EMF

Concept behind this question

More previous year questions on Motional EMF

Practice more Physics