Question

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as $B={{B}_{0}}{{e}^{\dfrac{-t}{\tau }}}$, where ${{B}_{0}},\tau $ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long time is,
$\begin{align}
  & a)\dfrac{{{\pi }^{2}}{{r}^{4}}{{B}_{0}}^{4}}{2\tau R} \\
 & b)\dfrac{{{\pi }^{2}}{{r}^{4}}{{B}_{0}}^{2}}{2\tau R} \\
 & c)\dfrac{{{\pi }^{2}}{{r}^{4}}{{B}_{0}}^{4}R}{2\tau } \\
 & d)none \\
\end{align}$

Answer 1

Hint: The magnetic field of the conducting material is varying along with time as given. We need to find the flux through the conducting material to find the heat generated. Heat generated is the integral of the emf of the circuit. Emf is the derivative of the flux through the circuit. Hence, we can find heat generated this way.
Formula used:
$\begin{align}
  & \phi =B.S \\
 & Q=\int\limits_{0}^{\infty }{\dfrac{{{\varepsilon }^{2}}}{R}dt} \\
\end{align}$

Complete step by step answer:
The given setup in the question is as follows:

let us first find the electric flux of the circuit,
$\begin{align}
  & \phi =B.S \\
 & \Rightarrow \phi ={{B}_{0}}{{e}^{\dfrac{-t}{\tau }}}\times \pi {{R}^{2}} \\
 & \\
\end{align}$
Now, the emf of the conducting material will be,
$\begin{align}
  & \varepsilon =-\dfrac{d\phi }{dt} \\
 & \Rightarrow \varepsilon =\dfrac{{{B}_{0}}\pi {{R}^{2}}{{e}^{\dfrac{-t}{\tau }}}}{\tau } \\
\end{align}$
Therefore, the heat generated in the conducting material will be,
$\begin{align}
  & Q=\int\limits_{0}^{\infty }{\dfrac{{{\varepsilon }^{2}}}{R}dt} \\
 & \Rightarrow Q=\int\limits_{0}^{\infty }{\dfrac{{{B}_{0}}^{2}{{\pi }^{2}}{{r}^{4}}{{e}^{\dfrac{-2t}{\tau }}}}{R{{\tau }^{2}}}} \\
 & \Rightarrow Q=\dfrac{{{\pi }^{2}}{{r}^{4}}{{B}_{0}}^{2}}{2B\tau } \\
\end{align}$
Therefore, the correct option is option b.

Additional information:
Electric flux is the rate of flow of the electric field through a given area. Electric flux is proportional to the number of electric field lines going through a virtual surface. Is the electric field being uniform throughout the material, the electric flux passing through a surface of vector area S please dot product between electric field and the surface? For a non-uniform electric field, dielectric flux through a small surface of area dS taken and then integrated throughout the whole area. Gauss’s law describes the electric flux or a surface SD as a surface integral. It is important that electric flux is not affected by charges that are not within the close to the surface where the S electric field gets affected by charges outside the closed surface.

Note:
The electric flux only gets affected by charges that are present inside the closed surface. If we consider the electric field, the electric field gets affected by the charges outside the closed surface too. The formula of electric flux is different for surfaces with uniform and non-uniform electric fields.