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A magnetic field directed into the page changes with time according to the expression B = (0.03t^2+1.4)T , where t is in seconds. The field has a circular cross-section of radius R = 2.5cm. What is the magnitude and direction of electric field at P, when t=3.0s 1 and r = 0.02 m. |
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Answer» Solution :`E =oint E.dl = (+dphi)/(dt)` `E(2pir) = A . (dB)/(dt) = PI r^2 xx (d)/(dt)(0.03 t^2 + 1.4)` ` E = (pi r^2)/(2pi r) xx (0.06t) = r/2 (0.06t)` `|E| = (0.02)/(2) xx 0.06 xx 3 = 18 xx 10^(-4) N//C` |
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