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A very long staright conductor and an isoceles triangular conductor lie in a plane and separated from each other as shown in the figure. Given a=10cm, b=20cm and h=10cm (a) Find their coefficeint of mutual induction. (b)If current in the staright wire is increasing at a rate of 2A//s , find the direction and magnitude of current in the triangular wire. [Diameter of the wire crosss-section d=1 mm, resistivity of the wire, rho=1.8xx10^(-8)Omega-m] |
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Answer» Solution :`(a)` Assuming a small strip at a distance `X` from the apex. `dA=(b)/(h)XDX` Let `I` be the current folowing in the straight wire, then magnetic field at the location of the strip is `B=(mu_(0)I)/(2pi(a+x))` `dphi=BdA=(mu_(0)IB)/(2pih)((xdx)/(a+x))` `phi=(mu_(0)Ib)/(2pih)int_(0)^(h)[1-(a)/(a+x)]DX` `phi=(mu_(0)Ib)/(2pih)=[h-a ln|(a+h)/(a)|]` The COEFFICENT of mutual induction is given by `M=(phi)/(I)=(mu_(0)b)/(2pih)[h-aln|(a+h)/(a)|]` Here, `b=20cm`, `h=10cm`, `mu_(0)=4pixx10^(-7)H//m` `M=((4pixx10^(-7))(0.2))/(2pi(0.1))[0.1-0.1ln|(10+10)/(10)|]=1.22xx10^(-8)H` `(b)` According to Faraday's law `E_("ind")=-M(di)/(dt)` since, `(di)/(dt)=2A//s` `:. (E_("ind")=(1.22xx10^(-8))(2)=2.44xx10^(-8)` volt. Resistance of the triangular conductor is `R=rho(1)/(A)` here, `rhp=1.8xx10^(-8)Omega-m`, `1=10sqrt(2)+20+10sqrt(2)=20(sqrt(2)+1)cm=0.48cm` `A=(pi)/(4)(10^(-3))^(2)=10^(-6)(pi)/(4)m^(2)` `:. R=(1.8xx10^(-8))((0.48))/((3.14))xx(4)/(10^(-6))=1.1xx10^(-2)Omega` `I=(E_("ind"))/(R )=(2.44xx10^(-8))/(1.1xx10^(-2))=2.2muA` The direction of the current is anticlockwise by Lenz Law. 
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