Problem Statement:
A rectangular loop with sides 10cm and 3cm is moving out of a region of uniform magnetic field of 0.5T directed normal to the loop. If we want to move the loop with a constant velocity of 1cm/sec, then what is the required mechanical force? Given resistance (R) is 1mΩ.

Solution:

Step 1: Calculate the magnetic flux through the loop.
The magnetic flux through a loop is given by the formula:
Φ = B * A * cos(θ)
where Φ is the magnetic flux, B is the magnetic field, A is the area of the loop, and θ is the angle between the magnetic field and the normal to the loop.

In this case, the magnetic field is 0.5T and the area of the loop is 10cm * 3cm = 30cm² = 0.003m². Since the magnetic field is normal to the loop, the angle θ is 0°, so cos(θ) = 1.

Therefore, the magnetic flux through the loop is:
Φ = 0.5T * 0.003m² * 1 = 0.0015Wb

Step 2: Calculate the induced emf in the loop.
The induced emf in a loop is given by Faraday's law of electromagnetic induction:
ε = -dΦ/dt
where ε is the induced emf and dΦ/dt is the rate of change of magnetic flux through the loop.

Since the loop is moving with a constant velocity of 1cm/sec, the rate of change of magnetic flux is zero. Therefore, the induced emf in the loop is zero.

Step 3: Calculate the required mechanical force.
The mechanical force required to move the loop with a constant velocity can be calculated using the equation:
F = ε / R
where F is the mechanical force, ε is the induced emf, and R is the resistance.

Since the induced emf is zero, the mechanical force required to move the loop with a constant velocity is also zero.

Conclusion:
The required mechanical force to move the loop with a constant velocity of 1cm/sec is zero. This is because the induced emf in the loop is zero, due to the loop moving with a constant velocity and the magnetic field being normal to the loop. Therefore, no mechanical force is needed to maintain the constant velocity of the loop.