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.

Explanation

`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`