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 is t -0-1 sec isA- 10 AB- 0-24 AC- 0-12 AD- 4-8 A

The correct option is B 0.24A ϕ =8t^2 4t+1 We have ϵ = dϕ/dt = 16t+4 ϵ l_t=0.1s= 1.6+4 =2.4 V i=ϵ/R =2.4/10 = 2.4/10 = 0.24 V ...

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Byju's Answer

Standard XII

Physics

Rotational EMF

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 $t^{2}$ -4t+1.The current induced in the coil at is t=0.1 sec is

10A

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0.24A

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0.12A

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4.8A

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Solution

The correct option is B
0.24A

$ϕ = 8 t^{2} - 4 t + 1$

We have

$ϵ = - \frac{d ϕ}{d t} = - 16 t + 4 ϵ l_{t = 0.1 s} = - 1.6 + 4 = 2.4 V i = \frac{ϵ}{R} = \frac{2.4}{10}$

= $\frac{2.4}{10}$ = 0.24 V

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

The correct option is B 0.24A ϕ=8t2−4t+1 We have ϵ=−dϕdt=−16t+4ϵlt=0.1s=−1.6+4=2.4 Vi=ϵR=2.410 = 2.410 = 0.24 V

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

The correct option is B
0.24A

$ϕ = 8 t^{2} - 4 t + 1$

We have

$ϵ = - \frac{d ϕ}{d t} = - 16 t + 4 ϵ l_{t = 0.1 s} = - 1.6 + 4 = 2.4 V i = \frac{ϵ}{R} = \frac{2.4}{10}$

= $\frac{2.4}{10}$ = 0.24 V