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PhysicsEMI/ACFaraday law/Lenz lawEasy2 minPYQ_2020

PhysicsEasynumerical

A conducting circular loop of area $2.5 \times 10^{- 3} m^{2}$ and resistance $10 Ω$ is placed perpendicular to a uniform time varying magnetic field $B (t) = 0.6 \sin (50 π t) T$ . What is the net charge (in $mC$ ) flowing through the loop during $t = 0$ and $t = 10 ms$ ?

Answer:

1.50

Solution:

Net charge $= \frac{Δ ϕ}{R} = \frac{A (B_{2} - B_{1})}{R}$

$= (2.5 \times 10^{- 3}) (0.6) [\sin \frac{50 π (10)}{10^{3}} - \sin 0]$

$= 1.5 \times 10^{- 3} [\sin \frac{π}{2} - \sin 0]$

$= 1.5 \times 10^{- 3} C = 1.5 mC$

Stream:JEESubject:PhysicsTopic:EMI/ACSubtopic:Faraday law/Lenz law

⏱ 2mℹ️ Source: PYQ_2020

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