Question

Using necessary figure (circuit), obtain an expression for self-induced emf produced in a coil.Answer the Given question.

Answer 1

magnetic field , B at the centre of coil carrying current , i with radius , r and having number of turns , n
$\qquad\bf{B=\frac{\mu_0}{2}\frac{ni}{r}}$

we know magnetic flux , Φ = BA
so, $\Phi=\frac{\mu_0}{2}\frac{ni}{r}\pi r^2$

$\Phi=\frac{\mu_0\pi nir}{2}$

we know from Lenz's law,
$\qquad L=\frac{n\Phi}{i}$
so, produced emf if $\frac{di}{dt}$ = R,
$\mathfrak{E}=L\frac{di}{dt}$

$\implies\mathfrak{E}=\frac{\mu_0\pi nir}{2}R$