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A square loop ACDE of area 20 cm^2 res...

A square loop `ACDE` of area `20 cm^2 ` resistance `5Omega` is rotate in as magnetic field `B=2T` through `180^@`
(a) in `0.01S`
(b) in `0.02 s`
Find the magnitudes of averasge values of e i and `/_\q` in both the cases.

Text Solution

Verified by Experts

Let us take the area vector S perpendicular to plane of loop inwards. So initially, `dSuarruarrB` and when it is rotated by `180^(@),SuarrdarrB.`
Hence, initial flux passing through the loop,
`phi_(i)=BScos0^(@)=(2)(20xx10^(-4))(1)`
`=4xx10^(-3)Wb`
Flux passing through the loop when it is rotated by `180^(@)`,
`phi_(f)=BScos180^(@)`
`=(2)xx(20xx10^(-4))(-1)=-4xx10^(-3)Wb`
Therefore, change in flux,
`Deltaphi_(B)=phi_(f)-phi_(i)=-8xx10^(-8)Wb`
(i) Given at `Deltat=0.01s, R=5Omega`
Using the relation,
`|e|=|-(Deltaphi_(B))/(Deltat)|=(8xx10^(-3))/(0.01)=0.8V`
or induced current `i=(|e|)/(R)=(0.8)/(5)=0.16A`
and charge `Deltaq=iDeltat=0.16xx0.01=1.6xx10^(-3)C`
(ii) Similarly, given at `Deltat=0.02s, R=5Omega`
`therefore" "|e|=|-(Deltaphi_(B))/(Deltat)|=(8xx10^(-3))/(0.02)=0.4V`
`therefore" "i=(|e|)/(R)=(0.4)/(5)=0.08A`
and charge `Deltaq=iDeltat=(0.08)(0.02)=1.6xx10^(-3)C`
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