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 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.

To find the charge that passes through the circuit first we have to find the relation between instantaneous current and instantaneous magnetic flux linked with it. The emf induced can be obtained by differentiating the expression of magnetic flux linked w.r.t. t and then applying Ohm's law, we get I= E R =1 R d dt According to the problem electric current is given as a function of time. I(t)= dQ dt or dQ dt =1 R d dt Integrating the variable separately in the form of differential equation for finding the charge Q that passed in time t, we have Q(t_1)=Q(t_2)=1 R [ (t_1)- (t_2) ] (t_1)L_1 _0 2 _x^ L_2+x dx' x' I(t_1) [Refer to the Eq. (i) of answer no. 25] = _0L_1 2 I(t_1)in L_2+x x Therefore the magnitude of charge is Q=1 R [ (T)- (0)] Q= _0L_1 2 in L_2+x x [I(T)-I(0)] Now I(T) = 0 and I(0) = 1. Q= _0L_1 2 I_0 in ( L_2+x x ) This is the required expression.

A rectangular loop of wire ABCD is kept close to an infinitely lo…

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Question

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current $\text{I}(\text{t})=\text{I}_0\Big(1-\frac{\text{t}}{\text{T}}\Big)$ for 0 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.

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Answer

To find the charge that passes through the circuit first we have to find the relation between instantaneous current and instantaneous magnetic flux linked with it. The emf induced can be obtained by differentiating the expression of magnetic flux linked w.r.t. t and then applying Ohm's law, we get
$\text{I}=\frac{\text{E}}{\text{R}}=\frac{1}{\text{R}}\frac{\text{d}\phi}{\text{dt}}$
According to the problem electric current is given as a function of time.
$\text{I(t)}=\frac{\text{dQ}}{\text{dt}}\text{ or }\frac{\text{dQ}}{\text{dt}}=\frac{1}{\text{R}}\frac{\text{d}\phi}{\text{dt}}$
Integrating the variable separately in the form of differential equation for finding the charge Q that passed in time t, we have
$\text{Q}(\text{t}_1)=\text{Q}(\text{t}_2)=\frac{1}{\text{R}}\big[\phi(\text{t}_1)-\phi(\text{t}_2)\big]$
$\phi(\text{t}_1)\text{L}_1\frac{\mu_0}{2\pi}\int_\text{x}^{\text{L}_2+\text{x}}\frac{\text{dx}'}{\text{x}'}\text{I}(\text{t}_1)$ [Refer to the Eq. (i) of answer no. 25]
$=\frac{\mu_0\text{L}_1}{2\pi}\text{I}(\text{t}_1)\text{in}\frac{\text{L}_2+\text{x}}{\text{x}}$
Therefore the magnitude of charge is
$\text{Q}=\frac{1}{\text{R}}[\phi(\text{T})-\phi(0)]$
$\text{Q}=\frac{\mu_0\text{L}_1}{2\pi}\text{ in }\frac{\text{L}_2+\text{x}}{\text{x}}[\text{I}(\text{T})-\text{I}(0)]$
Now I(T) = 0 and I(0) = 1.
$\therefore\ \text{Q}=\frac{\mu_0\text{L}_1}{2\pi}\text{I}_0\text{ in }\bigg(\frac{\text{L}_2+\text{x}}{\text{x}}\bigg)$
This is the required expression.

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