Exams › JEE Advanced › Physics
Correct answer: 0.6
The equilateral triangle of side 2 m has its base AB tangent to (or outside) the cylindrical region of radius 1 m. The flux through the triangle equals the flux through the cylinder (radius 1 m) since the field only exists inside r=1 m. Flux = B * pi * (1)² = (6/pi)*sin(2t)*pi = 6*sin(2t). EMF_max = d(6*sin(2t))/dt|_max = 12 V. Each side of the triangle has resistance R_side = 10/3 ohm. The loop has total resistance 10 ohm. Circuit: EMF drives current I = 12/10 = 1.2 A (max). Now, voltage across AB = I * R_AB if current flows uniformly — but we need the potential difference between A and B. Since this is a simple series loop, V_AB = EMF * (R_AB / R_total) only if AB is in series with rest... Actually in a single loop: V_AB (terminal PD across AB) = EMF - I * R_(other two sides) = 12 - 1.2*(20/3) = 12 - 8 = 4 V. That is also too large. The issue is that the EMF is distributed (it's an induced EMF in a loop, not a localized source). For an induced EMF, the electric field is non-conservative. The potential drop across AB = (R_AB / R_total) * EMF = (10/3)/10 * 12 = 4 V. Hmm. But the answer should be 0.6 V based on options. Let me reconsider — perhaps only the area of the triangle inside the cylinder contributes, or the geometry means only a fraction of the flux links AB. For equilateral triangle side 2 m, the inscribed circle has radius r_in = (2/sqrt(3))*(1/2) = 1/sqrt(3) ≈ 0.577 m < 1 m, so the cylinder of radius 1 m extends OUTSIDE the triangle. This means the full triangle area is INSIDE the cylinder region, not the other way. Area of equilateral triangle = (sqrt(3)/4)*4 = sqrt(3) m² ≈ 1.732 m². Flux = B * A_triangle = (6/pi)*sin(2t)*sqrt(3). EMF_max = (6/pi)*2*sqrt(3) = (12*sqrt(3))/pi ≈ 6.62 V. That's still not giving clean numbers. Re-reading: radius of cylindrical region = 1 m, triangle side = 2 m, triangle is placed symmetrically. The circumradius of equilateral triangle of side 2 = 2/sqrt(3) ≈ 1.155 m > 1 m, so the cylinder is inside the triangle. Only the circular cross-section (area = pi * 1² = pi m²) contributes to flux. So Flux = B * pi = (6/pi)*sin(2t)*pi = 6*sin(2t) Wb. EMF = d(Phi)/dt = 12*cos(2t), max EMF = 12 V. Total resistance R = 10 ohm, current I_max = 12/10 = 1.2 A. Now for potential across AB: AB is one side. By symmetry and noting that the EMF is induced in the loop and the field region is inside (the cylinder is fully inside the triangle): the EMF is localized in the region within the cylinder. The portion of each side of the triangle near the cylinder contributes to the EMF. However, AB being the bottom side is farther from the center (distance = height - apothem). Actually for the potential drop calculation across a particular resistor in an induced-EMF loop, V_AB = I * R_AB = 1.2 * (10/3) = 4 V if AB carries current I. But answer is 0.6 V. Perhaps the problem means the loop has sides 2 m but triangle is placed with only part intersecting the field. Given the answer choices and typical JEE problem design, maybe EMF = 12*cos(2t)*pi*1²/something, or perhaps only 1/20 of the area is inside. This problem likely requires the actual figure to determine the overlap area. Given the clean answer 0.6 V and total current would be 0.6*3 = 1.8 A or similar, the problem is figure-dependent.