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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 and r= 0.02m |
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Answer» Solution :`e= oint E.dl =(+d PHI)/(dt)` `E (2pi R)= A.(dB)/(dt)= PI r^(2) xx (d)/(dt) (0.03t^(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`/columns |
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