Question 71: Electromagnetic Induction

Question

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

Accepted Answer

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:
$$ \Phi = B \cdot A \cdot \cos(	heta) $$
Where:
$ B $ is the magnetic field strength,
$ A $ is the area of the loop, and
$ 	heta $ 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:
$$ A = 	ext{length} 	imes 	ext{width} = 2.5 \mathrm{~m} 	imes 2 \mathrm{~m} = 5 \mathrm{~m}^2 $$
Given the angle $ 	heta = 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
$$ +1 \mathrm{~V} $$

Answer

$$-2 \mathrm{~V}$$

Answer

$$+2 \mathrm{~V}$$

Answer

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

Answer

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

Electromagnetic Induction

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

Choose the correct answer: