A square loop of side 12 cm with its sides parallel to X and Y axes is moved with a velocity of 8 cm s^–1 in the positive x-direction in an environment containing a magnetic field in the positive z-direction. The field is neither uniform in space nor constant in time. It has a gradient of 10^–3 T cm^–1 along the negative x-direction (i.e., it increases by 10^–3 T cm^–3 as one moves in negative x-direction), and it is decreasing in time at the rate of 10^–3Ts^–1. Determine the direction and magnitude of the induced current in the loop if its resistance is 4.5 m Ω.

Given, a square loop.
side of the loop = 12 cm
velocity with which the loop is moving = 8 cm s^–1 = 8 x 10–2 m/s
Area of the loop, A = (12 x 10^–2) = 144 x 10^–4 m²

Gradient in magnetic field, 

$$\begin{gathered} \frac{\mathrm{dB}}{\mathrm{dt}} = {10}^3\mathrm{T}/\sec \\ \frac{\mathrm{dB}}{\mathrm{dx}} = {10}^{-3}\mathrm{T}/\mathrm{cm} = {10}^{-1} \mathrm{T}/\mathrm{m} \end{gathered}$$ 

Induced e.m.f. due to change of magnetic field B with time t is,

$$\begin{gathered} {\mathrm{\epsilon }}_1 = \frac{\mathrm{dQ}}{\mathrm{dt}} \frac{\mathrm{dB}.\mathrm{A}}{\mathrm{dt}} \frac{\mathrm{A}.\mathrm{dB}}{\mathrm{dt}} \\ =144 \times {10}^{-4} \times {10}^{-3} \end{gathered}$$ 

${\mathrm{\epsilon }}_1 = 144 \times {10}^{-7}\mathrm{V}$                                ...(I) 

Induced e.m.f. due to change of magnetic field B. with distance (x) is,
                     $$\begin{gathered} {\mathrm{\epsilon }}_2 = \frac{\mathrm{dQ}}{\mathrm{dt}} \frac{\mathrm{d}}{\mathrm{dt}}\mathrm{BA} \mathrm{A}.\frac{\mathrm{dB}}{\mathrm{dt}} \\ = \mathrm{A}. \frac{\mathrm{dB}}{\mathrm{dx}}.\frac{\mathrm{dx}}{\mathrm{dt}} \\ = 144 \times {10}^{-4} \times {10}^{-1} \times 8 \times {10}^{-2} \end{gathered}$$

${\mathrm{\epsilon }}_2 = 1152 \times {10}^{-7}$ 

Total e.m.f         ${\mathrm{\epsilon }}_1+{\mathrm{\epsilon }}_2 = 144 \times {10}^{-7} + 1152 \times {10}^{-7}$
         $= 1296 \times {10}^{-7}\mathrm{V}$ 

       $\mathrm{\epsilon } = 129.6 \times {10}^{-6}\mathrm{V}$ 
Resistance of the loop, R= 4.5 mΩ

Therefore, 

Induced current
                       $$\begin{gathered} \mathrm{I}= \frac{\mathrm{\epsilon }}{\mathrm{R}} = \frac{129.6 \times {10}^{-6}}{4.5 \times {10}^{-3}} \\ \cong 2.9 \times {10}^{-2}\mathrm{A}. \end{gathered}$$