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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=0.02 m. |
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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 (0.06)=(r)/(2)(0.06t)` `|E|=(0.02)/(2) xx 0.06 xx 3=18 xx 10^(-4) N//C` |
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