The current in the inner coil is `I = 2t^(2)`. Find the heat developed in the outer coil between t =0 and t seconds. The resistance of the outer coil is R and take `b gt gt a.`

Explanation

Let the current I be in the outer coil. The field at centre `B = (mu_(0)^(I))`
The flux through the inner coil = `(mu_(0)Ipia^(2))/(2b)`
The induced emf produced in the outer coil
`epsilon = -(dphi)/(dt)`
`(mu_(0)oua^(2))/(2b)(d)/(dt)(2t^(2)) = (2mu_(0)pia^(2)t)/(b)`
Current induced in the outer coil = `(epsilon)/(R ) = (2mu_(0)pia^(2)t)/(bR)`
Heat developed in the outer coil = `int_(0)^(t)l^(2)"Rdt"= int_(0)^(t)(4mu_(0)^(2)pi^(2)a^(4)t^(2)Rdt)/(b^(2)R^(2))=(4mu_(0)^(2)pi^(2)a^(4))/(b^(2)R) (t^(3))/(3)`