The magnetic flux - in weber linked with a coil of resistance 10 - varies with time t in second as -8 t 2-4 t -1- The current induced in the coil at t-0-1 s isA- 10 AB- 0-24 AC- 0-12 AD- 4-8 A

The correct option is B 0.24 AE.m.f induced, ϵ nbsp; = d ϕdt = ddt(8t^2 4t+1) = [(2 × 8)t 4]= [16t 4] Here, t=0.1 s ∴ϵ = [16 × 0.1 4] = 4 1.6=2.4 V Curr ...

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Standard XII

Physics

Dimensional Analysis

The magnetic ...

Question

The magnetic flux $ϕ$ (in weber) linked with a coil of resistance $10 Ω$ varies with time $t$ (in second) as $ϕ = 8 t^{2} - 4 t + 1$ . The current induced in the coil at $t = 0.1 s$ is

10 A

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0.24 A

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0.12 A

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4.8 A

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Solution

The correct option is B $0.24 A$
E.m.f induced, $ϵ = \frac{- d ϕ}{d t} = \frac{- d}{d t} (8 t^{2} - 4 t + 1) = - [(2 \times 8) t - 4] = - [16 t - 4]$
Here, $t = 0.1 s$
$∴ ϵ = - [16 \times 0.1 - 4] = 4 - 1.6 = 2.4 V$
Current induced $I = \frac{ϵ}{R} = \frac{2.4}{10} = 0.24 A$

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The magnetic flux ϕ (in weber) linked with a coil of resistance 10 Ω varies with time t (in second) as ϕ=8t2−4t+1. The current induced in the coil at t=0.1 s is

The correct option is B 0.24 AE.m.f induced, ϵ=−dϕdt=−ddt(8t2−4t+1)=−[(2×8)t−4]=−[16t−4] Here, t=0.1 s ∴ϵ=−[16×0.1−4]=4−1.6=2.4 V Current induced I=ϵR=2.410=0.24 A

The magnetic flux $ϕ$ (in weber) linked with a coil of resistance $10 Ω$ varies with time $t$ (in second) as $ϕ = 8 t^{2} - 4 t + 1$ . The current induced in the coil at $t = 0.1 s$ is

The correct option is B $0.24 A$
E.m.f induced, $ϵ = \frac{- d ϕ}{d t} = \frac{- d}{d t} (8 t^{2} - 4 t + 1) = - [(2 \times 8) t - 4] = - [16 t - 4]$
Here, $t = 0.1 s$
$∴ ϵ = - [16 \times 0.1 - 4] = 4 - 1.6 = 2.4 V$
Current induced $I = \frac{ϵ}{R} = \frac{2.4}{10} = 0.24 A$