Side of the Square loop, s = 12cm = 0.12m

Area of the loop, A = s × s = 0.12 × 0.12 = 0.0144 $m2$

The velocity of the lop, $v=8cm{s}^{-1}=0.08m{s}^{-1}$

The gradient of the magnetic field along negative x-direction,

$$$\frac{dB}{dx}={10}^{-3}Tc{m}^{-1}={10}^{-1}T{m}^{-1}$

And, the rate of decrease of the magnetic field,

$\frac{dB}{dt}={10}^{-3}T{s}^{-1}$

Resistance, $R<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>4.50m\mathrm{\Omega \hspace{0.17em}}=\hspace{0.17em}4.5\times {\mathrm{}}^{}$${10}^{-3}$Ω

The rate of change of the magnetic flux due to the motion of the loop in a non-uniform magnetic field is given as:

$\begin{array}{l}\frac{\mathrm{d\varphi }}{\mathrm{dt}}=\mathrm{A}\times \frac{\mathrm{dB}}{\mathrm{dx}}\times \mathrm{v}\\ =144\times {10}^{-4}{\mathrm{m}}^2\times {10}^{-1}\times 0.08\\ =11.52\times {10}^{-5}{\mathrm{Tm}}^2{\mathrm{s}}^{-1}\end{array}$

Rate of change of the flux due to explicit time variation in field B is given as:

$$$\begin{array}{l}\frac{{\mathrm{d\varphi }}^{\mathrm{\prime }}}{\mathrm{dt}}=\mathrm{A}\times \frac{\mathrm{dB}}{\mathrm{dt}}\\ =144\times {10}^{-4}\times {10}^{-3}\\ =1.44\times {10}^{-5}{\mathrm{Tm}}^2{\mathrm{s}}^{-1}\end{array}$

Since the rate of change of the flux is the induced emf, the total induced emf in the loop can be calculated as:

Hence, the direction of the induced current is such that there is an increase in the flux through the loop along the positive z-direction.