JEE MAIN - Physics (2003 - No. 19)
When the current changes from $$ + 2A$$ to $$-2A$$ in $$0.05$$ second, an $$e.m.f.$$ of $$8$$ $$V$$ is inducted in a coil. The coefficient of self- induction of the coil is
$$0.2H$$
$$0.4H$$
$$0.8$$ $$H$$
$$0.1$$ $$H$$
Explanation
$$e = - {{\Delta \phi } \over {\Delta t}} = {{ - \Delta \left( {LI} \right)} \over {\Delta t}} = - L{{\Delta I} \over {\Delta t}}$$
$$\therefore$$ $$\left| e \right| = L{{\Delta I} \over {\Delta t}} \Rightarrow 8 = L \times {4 \over {0.05}}$$
$$ \Rightarrow L = {{8 \times 0.05} \over 4} = 0.1H$$
$$\therefore$$ $$\left| e \right| = L{{\Delta I} \over {\Delta t}} \Rightarrow 8 = L \times {4 \over {0.05}}$$
$$ \Rightarrow L = {{8 \times 0.05} \over 4} = 0.1H$$
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