Question

A rectangular loop of length 2.5 m and width 2 m is placed at 60° to a magnetic field of 4 T. The loop is removed from the field in 10 sec. The average emf induced in the loop during this time is:

• A. -2V
• B. \(+2V\)
• C. \(+ 1V\)
• D. – 1V

Answer 1

The Correct Option is C

To calculate the average electromotive force (emf) induced in the loop, we can use Faraday's Law of electromagnetic induction, which states that the induced emf in a closed loop is equal to the negative rate of change of magnetic flux through the loop. Mathematically, this is given by:

\(\text{emf} = -\frac{\Delta \Phi}{\Delta t}\) 

where \(\Delta \Phi\) is the change in magnetic flux and \(\Delta t\) is the change in time.

1. Calculate the initial magnetic flux (\(\Phi_i\)):
The magnetic flux \((\Phi)\) through the loop is given by:

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

where \(B = 4 \, \text{T}\) (magnetic field strength), \(A = \text{length} \times \text{width} = 2.5 \, \text{m} \times 2 \, \text{m} = 5 \, \text{m}^2\) (area of the loop), and \(\theta = 60^\circ\) (angle with the magnetic field).

Plugging in the values, we get:

\(\Phi_i = 4 \times 5 \times \cos(60^\circ) = 4 \times 5 \times 0.5 = 10 \, \text{Wb}\)

2. Calculate the final magnetic flux (\(\Phi_f\)):
When the loop is removed from the field, \(B = 0\), hence \(\Phi_f = 0\).

3. Calculate the change in magnetic flux (\(\Delta \Phi\)):

\(\Delta \Phi = \Phi_f - \Phi_i = 0 - 10 = -10 \, \text{Wb}\)

4. Calculate the average emf:
Given that \(\Delta t = 10 \, \text{sec}\), the average emf is:

\(\text{emf} = -\frac{-10}{10} = +1 \, \text{V}\)

Thus, the average emf induced in the loop during this time is \(+1 \, \text{V}\). Therefore, the correct answer is \(+1 \, \text{V}\).

Answer 2

The average emf induced in the loop is given by:

\[ \text{Average emf} = -\frac{\Delta \Phi}{\Delta t} = -\frac{0 - (4 \times (2.5 \times 2) \cos 60^\circ)}{10} \]

Calculating the flux change:

\[ \Delta \Phi = 4 \times (2.5 \times 2) \times \frac{1}{2} = 10 \text{ Wb} \]

Then,

\[ \text{Average emf} = -\frac{-10}{10} = +1 \text{ V} \]