A rectangular loop of length $2.5 \mathrm{~m}$ and width $2 \mathrm{~m}$ is placed at $60^{\circ}$ to a magnetic field of $4 \mathrm{~T}$. The loop is removed from the field in $10 \mathrm{~sec}$. The average emf induced in the loop during this time is

<p>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:</p> <p>$\varepsilon = -\frac{d\Phi}{dt}$</p> <p>Where $ \varepsilon $ is the induced emf, and $ \Phi $ is the magnetic flux.</p> <p>To find the magnetic flux $ \Phi $ through the rectangular loop, we use the formula:</p> <p>$\Phi = B \cdot A \cdot \cos(\theta)$</p> <p>Where:</p> <ul> <li>$ B $ is the magnetic field strength,</li> <li>$ A $ is the area of the loop, and</li> <li>$ \theta $ is the angle between the magnetic field lines and the normal (perpendicular) to the plane of the loop.</li> </ul> <p>The area $ A $ of the rectangular loop is:</p> <p>$$ A = \text{length} \times \text{width} = 2.5 \mathrm{~m} \times 2 \mathrm{~m} = 5 \mathrm{~m}^2 $$</p> <p>Given the angle $ \theta = 60^\circ $, we can calculate the initial magnetic flux $ \Phi_{initial} $:</p> <p>$\Phi_{initial} = B \cdot A \cdot \cos(60^\circ)$</p> <p>$\Phi_{initial} = 4 \mathrm{~T} \cdot 5 \mathrm{~m}^2 \cdot \cos(60^\circ)$</p> <p>$\Phi_{initial} = 4 \cdot 5 \cdot \frac{1}{2} = 10 \mathrm{~Wb}$</p> <p>(Since $ \cos(60^\circ) = \frac{1}{2} $)</p> <p>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:</p> <p>$$ \Delta\Phi = \Phi_{final} - \Phi_{initial} = 0 - 10 \mathrm{~Wb} = -10 \mathrm{~Wb} $$</p> <p>The loop is removed from the field in $ t = 10 $ seconds, so the rate of change of magnetic flux is:</p> <p>$$ \frac{d\Phi}{dt} = \frac{\Delta\Phi}{\Delta t} = \frac{-10 \mathrm{~Wb}}{10 \mathrm{~s}} = -1 \mathrm{~Wb/s} $$</p> <p>Now we can find the average induced emf $ \varepsilon $:</p> <p>$\varepsilon = -\frac{d\Phi}{dt}$</p> <p>$\varepsilon = -(-1 \mathrm{~Wb/s})$</p> <p>$\varepsilon = +1 \mathrm{~V}$</p> <p>Therefore, the average induced emf in the loop during this time is $ +1 \mathrm{~V} $. The correct answer is:</p> <p>Option C</p> <p>$+1 \mathrm{~V}$</p>