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The two circuits in the figure interact through small concentric circular loops of radius R and r (r 1 and R 2 respectively. Inductances of the circuits are only due to circular loops. Direction of current are shown in the fig. Given : R=1cm., r=1mm , pi 2 =10. u 0 /4(pi) = 10 -7 N/A 2 , , R 1 10 ohm R 2 =1000 ohm , e 0 = 500V 1. The Kirchoff's loop equn for both primary and secondary circuit is given by... 2. Because R 2 is very large in comparision with other circuit parameters so di 2 /dt can be negleted. Find i 1 (t) for given approximation. Use i 1 (0)=0 3. After a long time current in the primary coil i 1 =50 Amp and in the secondary coil i 2 =0.005 A.At this instant what is the magnetic energy stored in the coil system....


The two circuits in the figure interact through small concentric circular loops of radius R and r (r<1 and R2 respectively. Inductances of the circuits are only due to circular loops. Direction of current are shown in the fig.


Given : R=1cm., r=1mm , pi2=10. u0/4(pi) = 10-7N/A2 , , R1 10 ohm  R2 =1000 ohm  , e0 = 500V


1. The Kirchoff's loop equn for both primary and secondary circuit is given by...


2. Because R2 is very large in comparision with other circuit parameters so di2/dt can be negleted. Find i1(t) for given approximation. Use i1(0)=0


3. After a long time current in the primary coil i1=50 Amp and in the secondary coil i2=0.005 A.At this instant what is the magnetic energy stored in the coil system....


Grade:11

1 Answers

ROSHAN MUJEEB
askIITians Faculty 829 Points
one year ago
Magnetic field at centre due to smaller loop
B1=μ04π.2πi1r1.......(i)B1=μ04π.2πi1r1.......(i)
Due to Bigger loopB2=μ04π.2πi2r2.......(i)B2=μ04π.2πi2r2.......(i)So net magnetic field at centreB=B1−B2=μ04π×2π(i1r1−i2r2)B=B1-B2=μ04π×2π(i1r1-i2r2)
According to questionB=12×B1B=12×B1
μ02π.2π(i1r1−i2r2)=12×μ04π.2πi1r1μ02π.2π(i1r1-i2r2)=12×μ04π.2πi1r1
i1r1−i2r2=i12r1⇒i12r1=i2r2⇒i2r2=1{r2=2r1}

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