TS EAMCET 7 May 2018 Shift 1 Solved Paper

The figure below represents the one vertex of square loop at origin, For wire $AB$ , magnetic filed, $B_0(1+0)=B_0$ $x=0$ For wire CD, magnetic filed, $B_0 \left(1+\frac{y}{L}\right)$ $x=L$ For wire $BC,$ magnetic filed, $B_0\left(1+\frac{x}{L}\right)$ $y=L$ For wire AD, magnetic filed, $B_0$ $y=0$ The forces in the square loop, $F_1=\int\limits_{0}^{L} I B_0 = I B_0 L$ $F_2=\int\limits_{0}^{L} I B_0 \left(1+\frac{x}{L}\right) \, dx = \frac{3}{2} I B_0 L$ $F_3=\int\limits_{0}^{L} I B_0 \left(1+\frac{x}{L}\right) \, dx = \frac{3}{2} I B_0 L$ $F_4=\int\limits_{0}^{L} I B_0 \, dx = I B_0 L$ The force in $x$ direction, $F_x=F_3-F_1$ $F_x=\frac{3}{2} I B_0 L - I B_0 L$ $F_x=\frac{1}{2} I B_0 L$ Similarly, The force in y direction, $F_y=F_2-F_4$ $F_y=\frac{3}{2} I B_0 L - I B_0 L$ $F_y=\frac{1}{2} I B_0 L$ The net magnitude of force, $F=\sqrt{ \left( \frac{1}{2} I B_0 L \right)^2 + \left( \frac{1}{2} I B_0 L \right)^2 }$ $F=\frac{I B_0 L}{\sqrt{2}}$