JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

A conducting circular loop of area 1.0 m $^{2}$ is placed perpendicular to a magnetic field which varies as B = sin(100 t) Tesla. If the resistance of the loop is 100 $Ω$ , then the average thermal energy dissipated in the loop in one period is ____ J.

A
$\frac{π}{2}$
B
$2 π$
C
$π$
D
$π^{2}$

Solution

Flux: $ϕ = B A = sin (100 t)$ . Induced EMF: $ε = - \frac{d ϕ}{d t} = - 100 cos (100 t)$ .

Instantaneous power dissipated: $P (t) = \frac{ε ^{2}}{R} = \frac{1 0 ^{4} cos ^{2} ( 100 t )}{100} = 100 cos^{2} (100 t)$ W.

Period: $T = \frac{2 π}{100} = \frac{π}{50}$ s. Average of $cos^{2}$ over a full period is $1/2$ .

Energy in one period: $Q = \overset{ˉ}{P} \cdot T = 50 \cdot \frac{π}{50} = π$ J.

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