Question

A rectangular loop of wire ABCD is kept close to an infinitely long wire carrying a current I(t) = Io (1 – t/T) for 0 ≤ t ≤ 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.

Answer 1

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 (dphi)/(dt)`

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

`I(t) = (dQ)/(dt)` or `(dQ)/(dt) = 1/R (dphi)/(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[phi(t_1) - phi(t_2)]`

For magnetic flux in rectangle:

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

`Q(t) = (mu_0I(t))/(2pix^')`

If length of strip is L₁ so total flux on strip of length L₁ at distance x' is 

`Q(t) = (mu_0I(t))/(2pix^') L_1`

X' varies from x to (x + L₂) so total flux in strip

`phi(t) = mu_0/(2pi) L_1 int_x^(x + L_2) (dx)/x^' I(t) = (mu_0L_1)/(2pi) I(t_1) log_e  ((L_2 + x))/x`

The magnetic of charge is given on length L₁

`int_Q_1 dQ = 1/R int dphi` from (III)

`int_0^Q dQ = 1/R * (mu_0L_1)/(2pi) log_e ((L_2 + x)/x) int_0^1 I(t_1)`

`Q = (mu_0L_1)/(R2pi) log_e ((L_2 + x)/x)(I - 0) = (mu_0L_1I_1)/(2piR) log  ((L_2 + x)/x)`