Ncert Exercises & Solutions

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

Hint: The change in current results in a change in magnetic field through the loop.

The emf induced can be obtained b differentiating the expression of magnetic flux linked wrt t nad then applying Ohm's law

$Step 1 : I = \frac{ε}{R} = \frac{1}{R} \frac{dφ}{dt}$

$We know that electric current;$

$I (t) = \frac{dQ}{dt} or \frac{dQ}{dt} = \frac{1}{R} \frac{dφ}{dt}$

$Step 2 : Integrating the variables separately in the form of a differential equation for finding the charge$

$Q that passed in time t, we have;$

$Q (t_{1}) - Q (t_{2}) = \frac{1}{R} [ϕ (t_{1}) - ϕ (t_{2})]$

$ϕ (t_{1}) = L_{1} \frac{μ_{0}}{2 π} \int_{x}^{L_{2} + x} \frac{dx'}{x'} I (t_{1})$

$= \frac{μ_{0} L_{1}}{2 π} I (t_{1}) \ln (\frac{L_{2} + x}{x})$

$Step 3 : The magnitude of charge is given by,$

$= \frac{μ_{0} L_{1}}{2 π} In \frac{L_{2} + x}{x} [I_{0} + 0]$

$= \frac{μ_{0} L_{1}}{2 π} I_{1} \times In (\frac{L_{2} + x}{x})$

$This is the required expression .$

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