Question

Answer this please please

A magnetic flux through a coil is perpendicular to its plane varying according to the relation $\phi$ (5t^3+4t^2+2t-5) Weber calculate the induced current through the coil at t= 2 s, if the resistance of coil is 5ohm

Answer 1

Answer= 15.6 Ampere

step by step explanation :

Resistance of the coil = 5 ohm (given)

instantaneous magnetic flux linked with coil then,

$e = - \frac{d \phi}{dt} = - \frac{d}{dt} (5t^{3} + 4t^{2} + 2t - 5) \\ = [5(3t^{2}) + 4(2t) + 2(1) - 0] \\ or \: e= \: ( - 15t^{2} + 8t + 2) \: V </p><p> \\ Therefore \: \: induced \: e.m.f \: produced \: in \: \\ the \: coil \: at \: t= 2 s, </p><p>\\ e = -[ 15( {2})^{2} + 8(2)^{2} + 2] </p><p>= - (60 + 16 + 2) = - 78 \: V</p><p> \\ Hence, \: magnitude \: of \: induced \: current \: in \: the \: coil \: at \: t= 2s, \\</p><p> I = \frac{e}{R} = \frac{78}{5} = 15.6 \: A$