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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` and (b) in `0.02 s` Find the magnitudes of averasge values of e i and `/_q` in both the cases. |
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Answer» Let the take the area vator `S` perpendicular to plane of loop inwards. So initially `dS` parallel to `B` and when it rorated by `180^(@)`, `S` is anti parallel to `B`. Hence, initial flux passing through the loop, `phi_(i) = BS cos 0^(@) = (2)(20 xx 10^(-4))(1)` `= 4 xx 10^(-3) Wb` Flux passing througn the loop when it rotated by `180^(@), phi_(f) = BS cos 180^(@)` `= (2) (20 xx 10^(-4))(-1) = -4.0 xx 10^(-3) Wb` Therefore, change in flux, `Delta phi_(B) = phi_(f)-phi_(i)`, `= -8 xx 10^(-3) Wb` (a) Given `Delta t = 0.01 s, R = 5 Omega` , `:. |e| = |(Delta phi_(B))/(Delta t)|` `= (8 xx 10^(-3))/(0.01) = 0.8V` , or `i = (|e|)/(R ) = (0.8)/(5) = 0.16 A` and `Delta q = i Delta t = 0.16 xx 0.01`, `= 1.6 xx 10^(-3) C` (b) `Delta t = 0.02 s`, `:. |e| = |-(Delta phi_(B))/(Delta t)|` `= (8 xx 10^(-3))/(0.02)`, `= 0.4V`, `i = = (|e|)/(R ) = (0.4)/(5) = 0.08A` and `Delta q = i Delta t = (0.08)(0.02)`, `= 16 xx 10^(-3)C` |
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