The flux associated with a coil changes from 1 . 35 Wb to 0 . 79 Wb within 1 10 s . Then the charge which flows in the coil, if resistance of coil is 7 Ω is

As, I = E R = d ϕ R d t ⇒ I d t = d ϕ R Integrating, ∫ I d t = ∫ d ϕ R ⇒ q = ϕ R If coil contains N turns, then q = N ϕ R . If there is flux change Δ ϕ , then q = N Δ ϕ R . ⇒ q = 1 7 × 1.35 - 0.79 = 0.08 C

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PhysicsEMI/ACInduced Electric FieldEasy2 minPYQ_2020

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The flux associated with a coil changes from $1.35 Wb$ to $0.79 Wb$ within $\frac{1}{10} s$ . Then the charge which flows in the coil, if resistance of coil is $7 Ω$ is

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Solution:

As, $I = \frac{E}{R} = \frac{d ϕ}{R d t}$
$\Rightarrow I d t = \frac{d ϕ}{R}$
Integrating,
$\int I d t = \int \frac{d ϕ}{R}$
$\Rightarrow q = \frac{ϕ}{R}$
If coil contains $N$ turns, then $q = \frac{N ϕ}{R}$ .
If there is flux change $Δ ϕ$ , then $q = \frac{N Δ ϕ}{R}$ .
$\Rightarrow q = \frac{1}{7} \times (1.35 - 0.79) = 0.08 C$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:EMI/ACSubtopic:Induced Electric Field

⏱ 2mℹ️ Source: PYQ_2020

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