Q : A rectangular wire frame of length 0.2 m, is located at a distance of 5 × 10^{–2} m from a long straight wire carrying a current of 10 A as shown in the figure.The width of the frame = 0.05 m. The wire is in the plane of the rectangle. Find the magnetic flux through the rectangular circuit. If the current decays uniformly to 0 in 0.2 s, find the emf induced in the circuit.

Solution : A current, i = 10 A is flowing in the long straight wire. Consider a small rectangular strip (in the rectangular wire frame) of width dx at a distance x from the straight wire.The magnetic flux at the location of the strip

$\displaystyle B = \frac{\mu_0 I}{2 \pi x}$

The flux linked with the infinitesimally small rectangular strip = B x Area of Strip

$\displaystyle d\phi = \frac{\mu_0 I}{2 \pi x} \times l dx $

where l is the length of the rectangular wire circuit = 2 × 10^{–1} m

Hence, the total magnetic flux linked with the rectangular frame

$\displaystyle \phi = \int d\phi = \int_{r_1}^{r_2} \frac{\mu_0 I l }{2 \pi x} dx $

$\displaystyle \phi = \frac{\mu_0 I l }{2 \pi} [log_e x]_{r_1}^{r_2} $

$\displaystyle \phi = \frac{\mu_0 I l }{2 \pi} [log_e r_2 – log_e r_1] $

$\displaystyle \phi = \frac{\mu_0 I l }{2 \pi} log_e \frac{r_2}{r_1} $

Substituting values, we get

φ = 2 × 10^{–7} × 10 × 2 × 10^{–1} × log_{e}2

= 2.772 × 10^{–7} Wb

Induced EMF $\displaystyle | e | = \frac{d\phi}{dt}$

$\displaystyle | e | = \frac{\mu_0 l log_e (\frac{r_2}{r_1})}{2 \pi} \frac{dI}{dt}$

= (2 × 10 × 2 × 10 log_{e} 2)(10/0.2)

= 1.386 × 10^{–6} V = 1.386 μV