Question

A rectangular loop of wire $$ABCD$$ is kept close to an infinitely long wire carrying a current $$I(t) = I_{0}\left (1 - \dfrac {t}{T}\right )$$ for $$0\leq t\leq T$$ and $$I(0) = 0$$ for $$t > T$$ (Fig.). Find the total charge passing through a given point in the loop, in time $$T$$. The resistance of the loop is $$R$$.

Answer 1

If $$(t)$$ is instantaneous current then,

$$I(t) = \dfrac {1}{R} \dfrac {d\phi}{dt} .... (I)$$

If $$Q$$ is charge passing in time $$t$$

$$\therefore I(t) = \dfrac {dQ}{dt} .... (II)$$

From (I) and (II)

$$\dfrac {dQ}{dt} = \dfrac {1}{R} \cdot \dfrac {d\phi}{dt}$$

Or $$dQ = \dfrac {1}{R} . d\phi .... (III)$$

Integrating both sides,

$$\int_{Q_{1}}^{Q_{2}} dQ = \dfrac {1}{R} \int_{\phi_{1}}^{\phi_{2}} d\phi$$

$$Q_{2} (t) - Q_{1}(t) =\dfrac {1}{R} [\phi_{2}(t) - \phi_{1}(t)]$$

For magnetic flux in rectangle:

Magnetic flux due to current carrying conductor at a distance $$x$$'

$$Q(t) = \dfrac {\mu_{0}I(t)}{2\pi x'}$$

If length of strip is $$L_{1}$$ so total flux on strip of length $$L_{1}$$ at distance $$x'$$ is

$$Q(t) = \dfrac {\mu_{0}I(t)}{2\pi x'} L_{1}$$

$$X'$$ varies from $$x$$ to $$(x + L_{2})$$ so total flux in strip

$$\phi (t) = \dfrac {\mu_{0}}{2\pi} \int_{x}^{x - L_{2}} \dfrac {dx}{x'} I(t) = \dfrac {\mu_{0}L_{2}}{2\pi} . I(t_{1})\log_{e} \dfrac {(L_{2} + x)}{x}$$

The magnitude of charge is given on length $$L_{1}$$

$$\int_{Q_{1}} dQ = \dfrac {1}{R} \int d\phi$$ [from (III)]

$$\int_{0}^{Q} dQ = \dfrac {1}{R} \cdot \dfrac {\mu_{0}L_{1}}{2\pi} \log_{e} \left (\dfrac {L_{2} + x}{x}\right ) \int_{0}^{1} I(t_{1})$$

$$Q = \dfrac {\mu_{0}L_{1}}{R2\pi} \log_{e} \left (\dfrac {L_{2} + x}{x}\right ) (I - 0) = \dfrac {\mu_{0}L_{1}L_{1}}{2\pi R} \log \left (\dfrac {L_{2} + x}{x}\right )$$.