Final answer:
The work done in pulling a square loop of resistance 10Ω out of a uniform magnetic field of 40.0 T uniformly in 1.0 sec is calculated to be 1.0 × 10⁻³ J.
Step-by-step explanation:
To determine the work done in pulling the loop out of the magnetic field, we first need to calculate the magnetic flux (Φ) through the square loop. The magnetic flux is given by the product of the magnetic field (B) and the area (A) through which it passes: Φ = B × A. Since the loop is perpendicular to the field, the magnetic flux would just be 40 T × 25 cm² = 40 T × 0.0025 m² = 0.1 Wb (webers).
According to Faraday's Law of Induction, when the magnetic flux changes over time, an electromotive force (EMF) is induced which can be calculated using ΕMF = -dΦ/dt. However, this does not directly answer the question about work done. Instead, we need to use the fact that the induced EMF causes a current to flow in the loop, which does work against resistance.
The induced current (I) can be calculated using Ohm's law, where I = ΕMF / R. The energy (work) dissipated by this current in the resistance over a time (t) is given by W = I²Rt. Substituting our values, first we find the induced EMF over 1 second (since Φ goes to zero as we pull it out, thus dΦ/dt = -0.1 Wb/s), which is 0.1 V. Now, we calculate the current: I = 0.1 V / 10 Ω = 0.01 A. The work done is W = (0.01 A)² × 10 Ω × 1 s = 1.0 × 10⁻³ J.