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A conducting loop in the form of a half circle of r=10cm is placed in a magnetic field as shown. The strength of the magnetic field is given by the relation B=5t^(2)+3t+5 where B is in Tesla and t- the time in seconds. The resistance of the loop is 3Omega. An ideal cell of e.m.f. E = 1.5V is connected to the loop. Calculate the induced e.m.f. and net current in the loop at t = 15 sec. |
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Answer» Solution :`E=1.5V,t=15sec, R=3Omega,B=5t^(2)+3t+5` Flux, `phi_(B)=BA` By Faraday.s law, induced e.m.f. = `varepsilon=(dphi_(B))/(dt)=A(dB)/(dt)=(pir^(2))/(2)(10t+3)` `3.14xx(100xx10^(-4))/(2)[10xx15+3]=2.4V` The direction of induced e.m.f. and APPLIED e.m.f. are OPPOSITE. Hence the net CURRENT is, `I=(varepsilon-E)/(R)=(2.4-1.5)/(3)=(0.9)/(3)=0.3A` |
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