Respuesta :
Explanation:
Let N be the number of turns in a circular loop having a radius of r₀ and a resistance R. The magnetic field is given by :
[tex]B(t)=B_oe^{-\dfrac{t}{\tau}}[/tex]
The induced emf in the circular coil is given by :
[tex]\epsilon=\dfrac{-d\phi}{dt}[/tex]
Where
[tex]\phi=BA[/tex]
[tex]\epsilon=N\dfrac{-d(BA)}{dt}[/tex]
[tex]\epsilon=NAB_o\dfrac{-d(e^{-\dfrac{t}{\tau}})}{dt}[/tex]
[tex]\epsilon=\dfrac{NAB_o}{\tau}e^{-t/\tau}[/tex]
Power dissipated is given by :
[tex]P=\dfrac{\epsilon^2}{R}[/tex]
[tex]P=\dfrac{(\dfrac{NAB_o}{\tau}e^{-t/\tau})^2}{R}[/tex]
[tex]P=(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}[/tex]
Energy dissipated in the circuit is given by :
[tex]E=\int\limits^T_0 {P.dt}[/tex]
[tex]E=\int\limits^T_0 {(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}.dt}[/tex]
[tex]E=\dfrac{1}{R}(\dfrac{NAB_o}{\tau})^2\int\limits^T_0 {e^{-2t/\tau}.dt}[/tex]
[tex]E=\dfrac{(N\pi r_o^2B_o^2)^2}{2R}{\tau(e^{2t/\tau}-1)}{}[/tex]
Hence, this is the required solution.
