A varying magnetic flux linking a coil is given by $\phi=\mathrm{xt}^{2}$. If at a time $\mathrm{t}=

Question

A varying magnetic flux linking a coil is given by $\phi=\mathrm{xt}^{2}$. If at a time $\mathrm{t}=3 \mathrm{~s}$, the emf induced is 9 V the value of $x$ is\

Professor PC Thomas Classes · Answer

$$ \phi=\mathrm{xt}^{2} $$ $\mathrm{d} \phi / \mathrm{dt}=2 \mathrm{xt} \quad \mathrm{e}=\frac{-\mathrm{d} \phi}{\mathrm{dt}}=-2 \mathrm{xt}$\ $X=\frac{-e}{2 t}$\ At $\mathrm{t}=3 \mathrm{sec}, \mathrm{x}=\frac{-9}{2 	imes 3}$\ $=-1.5 \mathrm{wb} / \mathrm{s}^{2}$\