JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Electromagnetic Induction

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

3 0^{\circ}

from the horizontal position, the induced EMF is 15.4 mV. The radius of the loop is ______ mm. (Take

π = \frac{22}{7}

)

Answer: 14

Solution

A loop of area

A

rotating at angular speed

ω

in a field

B

develops

ε = B A ω sin θ

, where

θ

is measured from the position of maximum flux. At

θ = 3 0^{\circ}

sin θ = \frac{1}{2}

, so with

A = π r^{2}

ε = \frac{1}{2} B ω π r^{2} .

Solving for the radius,

r^{2} = \frac{2 ε}{B ω π} = \frac{2 \times 15.4 \times 1 0 ^{- 3}}{0.5 \times 100 \times \frac{22}{7}} = 1.96 \times 1 0^{- 4} m^{2},

r = 1.4 \times 1 0^{- 2} m

. The radius of the loop is

14 mm

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer