A square loop of area 25 cm2 has a resistance of 10 Ω. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be2.5×10−3 J1.0×10−3 J1.0×10−4 J5×10−3 J

The work done in pulling the loop out of the magnetic field is equal to the energy dissipated in the resistor. This energy can be calculated using Faraday's law of electromagnetic induction and Ohm's law.

Faraday's law states that the induced emf in a circuit is equal to the rate of change of magnetic flux through the circuit. The magnetic flux (Φ) through the loop is given by the product of the magnetic field strength (B), the area of the loop (A), and the cosine of the angle between the field and the normal to the loop (cosθ). In this case, the loop is perpendicular to the field, so cosθ = 1. Therefore, Φ = B*A.

The rate of change of flux (dΦ/dt) is then equal to the change in flux divided by the change in time. Since the loop is being pulled out of the field, the final flux is zero, so the change in flux is -BA. The change in time is given as 1.0 sec. Therefore, dΦ/dt = -BA / 1.0 sec = -40.0 T * 25 cm^2 / 1.0 sec = -1000 T*cm^2/sec.

The induced emf (ε) is then equal to -dΦ/dt, so ε = 1000 V.

Ohm's law states that the current (I) in a circuit is equal to the voltage (V) divided by the resistance (R). Therefore, I = ε / R = 1000 V / 10 Ω = 100 A.

The energy (W) dissipated in the resistor is then given by I^2Rt = (100 A)^2 * 10 Ω * 1.0 sec = 1.0×10^5 J.

However, the area is given in cm^2, so we need to convert it to m^2 by multiplying by (1 m/100 cm)^2 = 1/10,000. Therefore, the actual energy dissipated is 1.0×10^5 J / 10,000 = 1.0×10^1 J = 10 J.

None of the given options match this result, so there may be a mistake in the problem or in my calculations.