Question

A square wire frame of side $a$ is placed $b$ away from a long straight conductor carrying current $I$ . The frame has resistance $R$ and self-inductance $L$ . The frame is rotated by $180^\circ$ about $OO'$ as shown in figure. Find the electric charge flown through the frame.

(A) $\dfrac{{2{\mu _0}i{a^2}}}{{2\pi Rb}}$
(B) $\dfrac{{{\mu _0}i}}{{2\pi R}}{\log _e}\dfrac{{b + a}}{{b - a}}$
(C) $\dfrac{{{\mu _0}ia}}{{2\pi R}}{\log _e}\dfrac{{b + a}}{{b - a}}$
(D) None of these

Answer 1

We know that the faraday's law stated that a changing magnetic field produces an electric field. So charges that are free to move will cause an EMF and a current. Thus the magnetic statement of Faraday’s Law is that EMF induced in a loop is proportional to rate of change in flux.
$E = iR = \dfrac{{\Delta \phi }}{{\Delta t}}$
We also know that the current in a conductor is nothing but, charge flowing in it per unit time $i = \dfrac{q}{t}$
We also know that, $\phi = \int {B.dA}$ .

Formulas used: We will be using the formula $E = iR = \dfrac{{\Delta \phi }}{{\Delta t}}$ where $E$ is the EMF induced by current $i$ , flowing through a conductor with resistance $R$ and $\Delta \phi$ is the change in flux of the body, while $\Delta t$ is the change in time intervals. We will also be using, $\phi = \int {B.dA}$ where $B$ is the magnetic field intensity produced due to the flux and $dA$ is the change in area.

Complete Step by Step Solution
We know that by faraday’s Laws of electromagnetism, $E = iR = \dfrac{{\Delta \phi }}{{\Delta t}}$ we also know that $q = it$ , Thus the charge induced on a body can be given by, $\dfrac{{{Q_{ind}}}}{{\Delta t}}R = \dfrac{{\Delta \phi }}{{\Delta t}}$ .
$\Rightarrow {Q_{ind}} = \dfrac{{\Delta \phi }}{R}$ .

Consider a small strip of width $dx$ at the distance of $x$ from the current carrying wire.So let us find the change in flux by finding the final flux and initial flux.
$d\phi = B.dA$
Integrating the expression to find the initial and final fluxes of the thin strip
$\int {d\phi } = \int {B.dA}$
We know that $\int {d\phi } = \phi$ and $B = \dfrac{{{\mu _0}I}}{{2\pi x}}$ . Also the are of the strip will be $A = a \times dx$ .
Substituting the values be get,
$\phi = \int {\dfrac{{{\mu _0}I}}{{2\pi x}} \times adx}$
The limits can be specified to be, $\left( {b - a} \right)$ to $b$ .
$\phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}\int_{b - a}^b {\dfrac{{dx}}{x}} = \dfrac{{{\mu _0}Ia}}{{2\pi }}\left[ {{{\log }_e}b - {{\log }_e}(b - a)} \right]$
$\Rightarrow \phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left( {\dfrac{b}{{b - a}}} \right)$
Now, consider the final position to find flux for,

Finding flux using integration,
$\phi = \int {\dfrac{{{\mu _0}I}}{{2\pi x}} \times adx}$
but the limits will be $b$ to $b + a$ .
$\phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}\int_b^{b + a} {\dfrac{{dx}}{x}} = \dfrac{{{\mu _0}Ia}}{{2\pi }}\left[ {{{\log }_e}(b + a) - {{\log }_e}(b)} \right]$
$\Rightarrow \phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left( {\dfrac{{b + a}}{b}} \right)$
Now we have both the initial and final flux let us find the differences between them, $\Delta \phi = {\phi _f} - ( - {\phi _i}) = \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left( {\dfrac{{b + a}}{b}} \right) - \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left( {\dfrac{b}{{b - a}}} \right)$
Since $\phi = B.dA\cos \theta$ and here $\theta = 180^\circ$
$\Delta \phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left[ {\left( {\dfrac{{b + a}}{b}} \right) \times \left( {\dfrac{b}{{b - a}}} \right)} \right]$
$\Rightarrow \Delta \phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left( {\dfrac{{b + a}}{{b - a}}} \right)$
Since we know that ${Q_{ind}} = \dfrac{{\Delta \phi }}{R}$ ,
$Q = \dfrac{{{\mu _0}Ia}}{{2\pi R}}{\log _e}\left( {\dfrac{{b + a}}{{b - a}}} \right)$
Hence the correct answer is option C.

Note
The problem can also be solved without breaking the limits and taking , solving for, $\phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}\int_{b - a}^{b + a} {\dfrac{{dx}}{x}} = \dfrac{{{\mu _0}Ia}}{{2\pi }}\left[ {{{\log }_e}(b + a) - {{\log }_e}(b - a)} \right]$
$\Rightarrow \phi = \dfrac{{{\mu _0}Ia}}{{2\pi }}\left[ {{{\log }_e}(b + a) - {{\log }_e}(b - a)} \right] = \dfrac{{{\mu _0}Ia}}{{2\pi }}{\log _e}\left( {\dfrac{{b + a}}{{b - a}}} \right)$
Thus ${Q_{ind}}$ will be, $Q = \dfrac{{{\mu _0}Ia}}{{2\pi R}}{\log _e}\left[ {\dfrac{{b + a}}{{b - a}}} \right]$ .