(a) As the magnetic field will be variable with distance from long straight wire, so the flux through square loop can be calculated by integration. Let us assume a width ‘dr’ of the square loop at a distance ‘r’ from straight wire B = $\frac{\mu_0}{4\pi} \frac{2I}{r}$ or ϕ = B . Adr = $\frac{\mu_0}{4\pi} \frac{2I}{r}$ adr As ϕ = MI ⇒ MI = $\frac{\mu_0 I a}{2\pi} \log_e$ (1 + a/x) M = $\frac{\mu_0 a}{2\pi} \log_e$ (1 + a/x) Total flux associated with square loop ϕ = ∫ dϕ = $\frac{\mu_0}{4\pi} 2I a \int\limits_{x}^{x+a} \frac{dr}{r}$ or ϕ = $\frac{\mu_0}{4\pi}$ 2Ia $\left[ \log_e r \right]_{x}^{x+a}$ ϕ = $\frac{\mu_0}{4\pi} 2I a \left[ \log_e \frac{x+a}{x} \right]$ or ϕ = $\frac{\mu_0 I a}{2\pi} \log_e$ (1 + a/x) (b) The square loop is moving right with a constant speed v, the instantaneous flux can be taken as ϕ = $\frac{\mu_0 I a}{2\pi} \log_e$ (1 + a/x) Induced emf, ε = – $\frac{d\phi}{dt}$ = - $\frac{d\phi}{dx} \frac{dx}{dt}$ = - v $\frac{d\phi}{dx}$ ε = - $\frac{\mu_0 I a v}{2\pi} \frac{d(\log_e(1+a/x))}{dx}$ ⇒ ε = $-\frac{\mu_0 I a v}{2\pi} \frac{1}{1+a/x} \left[ -\frac{a}{x^2} \right]$ or ε = $\frac{\mu_0}{2\pi} \frac{a^2 v}{x(x+a)}$ I or ε = 2 × $10^{-7}$ $\frac{[0.1]^2 \times 10 \times 50}{0.2[0.2+0.1]}$ = 1.67 × $10^{-5}$ V