JEE Advanced 2016 Paper 2

When loop was entering (x < L)$\phi = BLx$$e = -\frac{d\phi}{dt} = -BL\frac{dx}{dt}$$\left|e\right| = BLV$$i = \frac{e}{R} = \frac{BLV}{R}(\text{ACW})$F = ilB(Left direction)$= \frac{B^2 L^2 V}{R}$ (in left direction)$\Rightarrow a = \frac{F}{m} = -\frac{B^2 L^2 V}{mR}\; a = V\frac{dV}{dx}$$V\frac{dV}{dx} = -\frac{B^2 L^2 V}{mR} \Rightarrow \int\limits_{V_0}^{V} dV = -\frac{B^2 L^2 V}{mR} \int\limits_{0}^{x} dx$$\Rightarrow V = V_0 - \frac{B^2 L^2}{mR}x$ (straight line of negative slope for x < L)$I = \frac{BL}{R}V \Rightarrow$(I vs x will also be straight line of negative slope for x < L)$L \leq x \leq 3L$$\frac{d\phi}{dt} = 0$e = 0 i = 0F = 0x > 4Le = BlVForce also will be in left direction$i = \frac{BLV}{R}$ (clock wise) $a = -\frac{B^2 L^2 V}{mR} = V\frac{dV}{dx}$$F = \frac{B^2 L^2 V}{R}$       $\int\limits_{L}^{x} -\frac{B^2 l^2}{mR} = \int\limits_{v_i}^{v_f} dV$$\Rightarrow -\frac{B^2 L^2}{mR}(x-L) = V_f - V_i$$V_f = V_i - \frac{B^2 L^2}{mr}(x-L)$ (straight line of negative slope)$I = \frac{BLV}{R} \rightarrow$(Clockwise) (straight line of negative slope)