According to Faraday's Law of Electromagnetic Induction, the induced emf in a closed circuit is equal to the negative rate of change of magnetic flux through the circuit. Mathematically, it is expressed as:

$$ \varepsilon = -\frac{d\Phi}{dt} $$

Where $ \varepsilon $ is the induced emf, and $ \Phi $ is the magnetic flux.

To find the magnetic flux $ \Phi $ through the rectangular loop, we use the formula:

$$ \Phi = B \cdot A \cdot \cos(\theta) $$

Where:

• $ B $ is the magnetic field strength,
• $ A $ is the area of the loop, and
• $ \theta $ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.

The area $ A $ of the rectangular loop is:

$$ A = \text{length} \times \text{width} = 2.5 \mathrm{~m} \times 2 \mathrm{~m} = 5 \mathrm{~m}^2 $$

Given the angle $ \theta = 60^\circ $, we can calculate the initial magnetic flux $ \Phi_{initial} $:

$$ \Phi_{initial} = B \cdot A \cdot \cos(60^\circ) $$

$$ \Phi_{initial} = 4 \mathrm{~T} \cdot 5 \mathrm{~m}^2 \cdot \cos(60^\circ) $$

$$ \Phi_{initial} = 4 \cdot 5 \cdot \frac{1}{2} = 10 \mathrm{~Wb} $$

(Since $ \cos(60^\circ) = \frac{1}{2} $)

When the loop is removed from the magnetic field, the final magnetic flux $ \Phi_{final} $ is zero, because the loop is no longer within the magnetic field. Thus, the change in magnetic flux $ \Delta\Phi $ is:

$$ \Delta\Phi = \Phi_{final} - \Phi_{initial} = 0 - 10 \mathrm{~Wb} = -10 \mathrm{~Wb} $$

The loop is removed from the field in $ t = 10 $ seconds, so the rate of change of magnetic flux is:

$$ \frac{d\Phi}{dt} = \frac{\Delta\Phi}{\Delta t} = \frac{-10 \mathrm{~Wb}}{10 \mathrm{~s}} = -1 \mathrm{~Wb/s} $$

Now we can find the average induced emf $ \varepsilon $:

$$ \varepsilon = -\frac{d\Phi}{dt} $$

$$ \varepsilon = -(-1 \mathrm{~Wb/s}) $$

$$ \varepsilon = +1 \mathrm{~V} $$

Therefore, the average induced emf in the loop during this time is $ +1 \mathrm{~V} $. The correct answer is:

Option C

$$ +1 \mathrm{~V} $$