### Related Formula Motional electromotive force formula: epsilon = B cdot v cdot l_texteff $\epsilon = B \cdot v \cdot l_{\text{eff}} $ where l_texteff$l_{\text{eff}}$ is the net straight-line displacement distance joining start point A$A$ to target endpoint E$E$ (l_AE$l_{AE}$). ### Core Logic Since the external magnetic induction field map remains completely uniform across space, any arbitrary zig-zag wire loop profile can be simplified into a straight line connector vector bridging its raw boundary tips.

### Step 1: Compute Effective Length From the geometric angle configurations: l_texteff = 2 cdot (10 cdot sin 45^circ) = 2 cdot 10 cdot frac1sqrt2 = 10sqrt2mathrm~cm = 0.1sqrt2mathrm~m $ l_{\text{eff}} = 2 \cdot (10 \cdot \sin 45^{\circ}) = 2 \cdot 10 \cdot \frac{1}{\sqrt{2}} = 10\sqrt{2}\mathrm{~cm} = 0.1\sqrt{2}\mathrm{~m} $ (Alternatively tracking structural vector projections: l_AB = 2 cdot 10sin(45^circ)mathrm~cm$l_{AB} = 2 \cdot 10\sin(45^{\circ})\mathrm{~cm}$). ### Step 2: Calculate Induced EMF Substitute the parameters into the induction formula: epsilon = left( frac1sqrt2 ight) times left( 10 times 10^-2mathrm~m/s ight) times left( 10sqrt2 times 10^-2mathrm~m ight) $ epsilon = \left( \frac{1}{\sqrt{2}} ight) \times \left( 10 \times 10^{-2}\mathrm{~m/s} ight) \times \left( 10\sqrt{2} \times 10^{-2}\mathrm{~m} ight) $ epsilon = frac1sqrt2 cdot (0.1) cdot (0.1sqrt2) = 0.01mathrm~V = 10mathrm~mV$ epsilon = \frac{1}{\sqrt{2}} \cdot (0.1) \cdot (0.1\sqrt{2}) = 0.01\mathrm{~V} = 10\mathrm{~mV}$ ### Pattern Recognition In uniform magnetic fields, intermediate paths do not affect motional EMF values. The output voltage depends strictly on the straight line distance joining the end tips perpendicular to the velocity direction. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Physics: Electromagnetic Induction