Question

What is the energy dissipated as a function of time in a circular loop of N turns of wire having a radius of r0 and a resistance of R if the plane of the loop is perpendicular to a magnetic field given by B(t)=B0e−t/τ?

Answer 1

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 :

$B(t)=B_oe^{-\dfrac{t}{\tau}}$

The induced emf in the circular coil is given by :

$\epsilon=\dfrac{-d\phi}{dt}$

Where

$\phi=BA$

$\epsilon=N\dfrac{-d(BA)}{dt}$

$\epsilon=NAB_o\dfrac{-d(e^{-\dfrac{t}{\tau}})}{dt}$

$\epsilon=\dfrac{NAB_o}{\tau}e^{-t/\tau}$

Power dissipated is given by :

$P=\dfrac{\epsilon^2}{R}$

$P=\dfrac{(\dfrac{NAB_o}{\tau}e^{-t/\tau})^2}{R}$

$P=(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}$

Energy dissipated in the circuit is given by :

$E=\int\limits^T_0 {P.dt}$

$E=\int\limits^T_0 {(\dfrac{NAB_o}{\tau})^2\dfrac{e^{-2t/\tau}}{R}.dt}$

$E=\dfrac{1}{R}(\dfrac{NAB_o}{\tau})^2\int\limits^T_0 {e^{-2t/\tau}.dt}$

$E=\dfrac{(N\pi r_o^2B_o^2)^2}{2R}{\tau(e^{2t/\tau}-1)}{}$

Hence, this is the required solution.