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following is the given circuit which contains two resistance R1 and R2 in form of circle of radius r=1m with a battery having e.m.f=10pi volt.Upper resistance is having resistivity p1=4ohm m and lower resistance having resistivity p2=2 ohm m . Angle between two points Aand B is 60.(wires have same area cross section A1=A2=2 cm^2

Ayush Gupta , 8 Years ago
Grade 10
anser 1 Answers
ROSHAN MUJEEB
R1 =4(2)(1/6)/0.0002

R2 =2(2)(5/6) /0.0002

R1:R2=2:5

current in upper arm (iupper) =10 /[4(2)(1/6)/0.0002] =.0015Ampere

current in lower arm (ilower)= 10/[2(2)(5/6) /0.0002]=.0006Ampere

net curren t = .0015 +.0006= 0.0021 Ampere =21(10)-4

μ0=1.26(10)-6

B= (μ0ilowerΦ2/4) - (μ0iupperΦ1/4)

Φ1= /3

Φ2= 5/3

put these value and find B
Last Activity: 5 Years ago
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