The flux of magnetic field through a closed conducting loop of resistance `0.4 Omega` changes with time according to the equation `Phi =0.20 t^(2)+0.4

Question

The flux of magnetic field through a closed conducting loop of resistance `0.4 Omega` changes with time according to the equation `Phi =0.20 t^(2)+0.4

The flux of magnetic field through a closed conducting loop of resistance `0.4 Omega` changes with time according to the equation `Phi =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)charge passed through the loop in `t=0` to `t=5s` (iv)average current.In time interval `t=0` to `t=5 s` (v) heat produced in `t=0` to `t=5s`

Monjur Alahi

5 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - A::B::C::D
`epsilon=-(dphi)/(dt)=-(0.4 t+0.4)`
(i)`epsilon_(t=2)=-1.2` volt
(ii)`ltepsilongt = (Deltaphi)/(Delta t)=([0.2(5)^(2)+0.4(5)+0.6]-[0.6])/(5-0)=1.4` volt
(iii)`Deltaq=(Deltaphi)/R=17.5 C.` (iv)`lt I gt =(Deltaq)/(Deltat)=3.5` Anticlockwise
(v) `H=int epsilon^(2)/R.dt=underset(0)overset(5)int([0.4t+0.4]^(2))/R.dt=86/3J`