Question

The flux of magnetic field through a closed conducting loop of resistance 0.4$\mho$ changes with time according to the equation $\varphi$ = 0.20$t^2$ + 0.40t + 0.60 where t is time in seconds. Find (i) the induced emf at t = 2s. (ii) the average induced emf in t = 0 to t = 5 s. (iii) change passed through the loop in t = 0 to t = 5s. (iv) average current in time interval t = 0 to t = 5s (v) heat produced in t = 0 to t = 5s.

Answer 1

i) $\mid E\mid =\frac{d\varphi }{dt}=0.4t+0.4=0.8+0.4=1.2Volt$

ii) $\mid E_{av}\mid =\frac{\Delta \varphi }{\Delta t}=\frac{{\varphi }_5-{\varphi }_0}{5-0}=\frac{\left[\right(0.2\times 25+0.4\times 5+0.6)-0.6]}{5}=1.4Volts$

iii) As $E=IR$

or $I=\frac{E}{R}$

$\frac{dq}{dt}=\frac{E}{R}$

$dq=\frac{E}{R}dt$

${\int }_0^{Q}dq=\frac{1}{R}{\int }_0^5(0.4t+0.4)dt$

$Q=\frac{1}{0.4}[0.2t^2+0.4t{]}_0^5=\frac{(0.2\times 25)+(0.4\times 5)}{0.4}=17.5Columbs$

iv) $I_{avg}=\frac{\Delta Q}{\Delta t}=\frac{17.5}{5}=3.5A$

v) As $dH=I^{2}Rdt=\frac{V^2}{R}dt$

As $V=0.4t+0.4$

$\therefore {\int }_0^{H}dH={\int }_0^5\frac{(0.4t+0.4{)}^2}{0.4}dt$

$H=0.4[\frac{t^3}{3}+t^2+t{]}_0^5=\frac{86}{3}Joule$