Question

The area of an elastic circular loop decreases at a constant rate, dA/dt=−5.00×10⁻³m²/s. The loop is in a magnetic field B=0.40 T whose direction is perpendicular to the plane of the loop. At t=0, the loop has area A=0.285 m². Determine the induced emf at t=0, and at t=2.00 s.

Answer 1

The induced emf at t = 0 is 3.50 x 10^-2 V and at t = 2.00 s is -7.00 x 10^-2 V in a circular elastic loop with a constant decrease in area.

Step-by-step explanation:

The induced emf in a circular loop is given by the equation E = -dA/dt, where E is the induced emf and dA/dt is the rate of change of area with time. In this case, dA/dt is given as -3.50 x 10^-2 m^2/s. To determine the induced emf at t = 0, we substitute the given values into the equation and calculate:

E = -(-3.50 x 10^-2 m^2/s) = 3.50 x 10^-2 V

To determine the induced emf at t = 2.00 s, we multiply the rate of change of area by the time:

E = -(3.50 x 10^-2 m^2/s) x (2.00 s) = -7.00 x 10^-2 V

Therefore, the induced emf at t = 0 is 3.50 x 10^-2 V, and the induced emf at t = 2.00 s is -7.00 x 10^-2 V.