Question

The magnetic flux through a coil is $4\times {10}^{-4}W/b/m^2$ at time $t=0$.It reduces to $10\%$ of its original value in 't' seconds.If the induced e.m.f is $0.72mV,$ then the time t is:

• A. $0.25s$
• B. $0.05s$
• C. $0.75s$
• D. $1s$

Answer 1

The correct option is B $0.05s$

As given in question

magnetic turn $=\left({\varphi }_1\right)=4\times {10}^{-4}wb/m^2$

at $t=0$

At, $t=t_2$, the turn reduces to $10\%$, means

${\varphi }_2=0.9{\varphi }_1$

As, per the farady's law,

In duced Emf $=\frac{Nd\varphi }{dt}$

$e=\frac{nd\varphi }{dt}$ ...(1)

$e=0.72mu$ pur in (1), take $N=1$ turns are constant

$0.72\times {10}^{-3}=\frac{-({\varphi }_2-{\varphi }_1)}{(t_2-t_1)}$

$0.72\times {10}^{-3}=\frac{-(0.9-2){\varphi }_1}{(t_2-0)}$

$t_2=\frac{\left(0.1\right)\times (4\times {10}^{-4})}{(0.72\times {10}^{-3})}$

$t_2=0.05sec$