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  • A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t)=I_{0}(I-\frac{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.

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t)=I_{0}(I-\frac{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.

Answers (1)

I=\frac{E}{R}=\frac{1}{R}\frac{d\phi }{dt}

According to the problem electric current is given as a function of time.

I(t)=\frac{dQ}{dt}\; or\; \frac{dQ}{dt}=\frac{1}{R}\frac{d\phi }{dt}

Integrating the variable separately in the form of the differential equation for finding the charge Q that passed in time t, we have

Q(t_{1})-Q(t_{2})=\frac{1}{R}\left [ \phi (t_{1})-\phi (t_{2}) \right ]

Q(t_{1})=L_{1}\frac{\mu _{0}}{2\pi}\int_{x}^{L_{2}+x}\frac{dx'}{x'}I(t_{1}) [Refer to the Eq. (i) of answer no.25]

=\frac{\mu _{0}L_{1}}{2\pi}I(t_{1})ln\frac{L_{2}+x}{x}

Therefore the magnitude of the charge is

Q=\frac{1}{R}\left [ \phi (T)-\phi (0) \right ]

=\frac{\mu _{0}L_{1}}{2\pi}I(t_{1})ln\frac{L_{2}+x}{x}[I(T)-I(0)]

Now I(T)=0\; and \; I(0)=1

\therefore Q=\frac{\mu _{0}L_{1}}{2\pi}I_{0}In \left ( \frac{L_{2}+x}{x} \right )

 

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infoexpert23

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