Exams › JEE Advanced › Physics

Three resistors, each of resistance R, are connected in the form of an equilateral triangle of side a. The assembly is placed in a time-varying magnetic field B = B₀ * e^(-lambda*t) directed perpendicular to the plane of the triangle. Find the magnitude of the induced current in the circuit.

1. (a² * lambda) / (2 * sqrt(3) * R) * B₀ * e^(-lambda*t)
2. (a² * lambda) / (4 * sqrt(3) * R) * B₀ * e^(-lambda*t)
3. (a² * B₀) / (4 * lambda * sqrt(3) * R) * e^(-lambda*t)
4. (a² * B₀ * R) / (4 * lambda * sqrt(3)) * e^(-lambda*t)

Correct answer: (a² * lambda) / (4 * sqrt(3) * R) * B₀ * e^(-lambda*t)

Solution

Magnetic flux: Phi = B * A = B₀ * e^(-lambda*t) * (sqrt(3)/4) * a². EMF = -d(Phi)/dt = B₀ * lambda * e^(-lambda*t) * (sqrt(3)/4) * a². The equilateral triangle circuit: all three R in a loop. For the induced EMF driving the loop, the total loop resistance = R + R + R = 3R? No — the current divides. The three resistors form a triangle. The EMF is induced in the loop. For current calculation: consider the triangle as a loop. The total resistance around the loop is 3R, but since each side is the same and the EMF is distributed (all sides are in the field), it's like an EMF in series with the total loop resistance 3R. Wait: actually, by symmetry, the induced current is the same in all three sides. EMF = (sqrt(3)/4) * a² * lambda * B₀ * e^(-lambda*t). Effective resistance = 3R (going around the loop). But when we tap across any one resistor, the other two (in series = 2R) are in parallel with R. For the loop current analysis: the induced EMF acts around the entire loop, so loop current i = EMF / (3R). But the question might consider the 'effective resistance' as 2R/3 (one R parallel with two R in series). Actually with uniform B over the triangle, it's a single loop with total resistance 3R. Current = EMF/(3R) = (sqrt(3)*a²*lambda*B₀*e^(-lambda*t)) / (4*3R) = (sqrt(3)*a²*lambda*B₀*e^(-lambda*t)) / (12R) = a²*lambda*B₀*e^(-lambda*t) / (4*sqrt(3)*R). This matches option (b).

Related JEE Advanced Physics questions

• What is the final velocity attained by the loop?
• A long coaxial cable is made up of two cylindrical conductors with radii a and b. The inner conductor carries a constant current i, while the outer conductor acts as the return path for the current. What is the self-inductance of a segment of the cable of length l?
• Determine the inductance per unit length for a loop created by connecting the ends of two infinitely long parallel wires, each with radius r, and separated by a distance d between their centers. Ignore the magnetic flux inside the wires and the effects at the ends.
• If the electromotive force is 10 V and the rate of change of current is calculated as 2.5 A/s, what is the self-inductance of the coil, given by L = e / (dl/dt)?
• The magnitude of the induced electric field in the orbit at any instant of time during the time interval of the magnetic field change is
• A semicircular loop of radius r is positioned on the axis of a large current-carrying circular ring of radius R (with R >> r). The loop is placed at an axial distance of R*sqrt(3) from the center of the ring, and the plane of the semicircular loop makes an angle of 53 deg with the plane of the ring. Find the mutual inductance M between the two coils.

⚔️ Practice JEE Advanced Physics free + battle 1v1 →