18. A magnetic field of $2 \times 10^{-2} T$ acts at right angles to a coil of area $100 cm^2$, with $50$ turns. The average e.m.f. induced in the coil is $0.1 V$, when it is removed from the field in $t sec$. The value of $t$ is (1991)
Solution:
Initial flux $\Phi_i = N B A = 50 \times (2 \times 10^{-2}) \times (100 \times 10^{-4}) = 10^{-2} Wb$.
Final flux $\Phi_f = 0$. Magnitude of average emf $e = \frac{\Delta \Phi}{t} = \frac{10^{-2}}{t}$.
Given $e = 0.1 V$, we have $0.1 = \frac{10^{-2}}{t}$, which gives $t = \frac{10^{-2}}{0.1} = 0.1 s$.
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