# 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 ….

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 × loge2

= 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 loge 2)(10/0.2)

= 1.386 × 10–6 V = 1.386 μV