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`]

Explanation

`(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.