Get Answers to all your Questions

header-bg qa

Magnetic flux linked with a closed circuit of resistance 10ohm varies with time t as 5t2-4t+1. the induced emf in the circuit at t=0.2sec.

Answers (1)

\\\text{Magnetic flux }\phi=5 \mathrm{t}^{2}-4 \mathrm{t}+1$ \\ Time $t=0.2$ sec \\ Now, the induced e. $\mathrm{m}$. $\mathrm{f}$ in the circuit $E=-\frac{d \phi}{d t}$ \\ $\mathrm{E}=-\frac{\mathrm{d}\left(5 \mathrm{t}^{2}-4 \mathrm{t}+1\right)}{\mathrm{dt}}$ \\ $\mathrm{E}=-10 \mathrm{t}+4$ \\ Now, at $t=0.2 \mathrm{s}$ \\ $\mathrm{E}=-10 \times 0.2+4$ \\ $\mathrm{E}=2 \mathrm{V}$ \\ Hence, the induced emf in the circuit is $2 \mathrm{V}$

Posted by

lovekush

View full answer