InterviewSolution
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Figure shows two circular regions `R_(1)` and `R_(2)` with redii `r_(1) = 21.2 cm` and `r_(2) = 32.3 cm`, respectively. In `R_(1)` there is a uniform magetic field `B_(1) = 48.6 mT` into the page and in `R_(2)` there is a uniform magnetic field `B_(2) = 77.2 mT` out of the page (ignore any fringing of these fields).Both fields are decreasing at the rate `8.50 mT s^(-1)`. Calculate the intergal `oint vec(E). vec(dl)` for each of the three identical paths. |
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Answer» `oint vec(E) . dvec(s) = - (d phi)/(dt) = - A (dB)/(dt)` For loop a: `oint vec(E) . vec(dl) = - (d(vec(B) . vec(A)))/(dt) = (d(BA cos 180^(@)))/(dt)` `= a(dB)/(dt) = pi r_(1)^(2) (-8.5 mT//s)` `= - pi (21.2 xx 10^(-2))^(2) (8.5 xx 10^(-3))` `= - 1.2 xx 10^(-3) V` For loop b: `oint vec(E) .vec(dl) = - (d(vec(B) .vec(A)))/(dt) = A (dB)/(dt)` `= - pi ((32.3)/(100))^(2) xx 8.5 xx 10^(-3) = - 1.2 xx 10^(-3) V` For loop c: `oint vec(E) .vec(dl) = - (d(vec(B) .vec(A)))/(dt) = - (d(vec(B_(1))*vec(A_(1)) + vec(B_(2))* vec(A_(2))))/(dt)` `= - (d(B_(1)A_(1) cos180^(@) + B_(@)A_(2) cos0^(@)))/(dt)` `= + A_(1) (dB_(1))/(dt) - A_(2) (dB_(2))/(dt)` `= - 1.2 xx 10^(-3) - (-2.7 xx 10^(-3))` `= - 1.5 xx 10^(-3) V` |
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