Question

A rectangular loop of length \( 2.5 \) m and width \( 2 \) m is placed at \( 60^\circ \) 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. \( -2 \) V
• B. \( +2 \) V
• C. \( +1 \) V
• D. \( -1 \) V

Hint:

The induced emf is found using Faraday’s Law, considering the flux change over time.

Answer 1

The Correct Option is C

Step 1: {Using Faraday’s Law}
\[ {Average emf}, \varepsilon = -\frac{\Delta \Phi}{\Delta t} \] Step 2: {Calculating the Magnetic Flux}
\[ \Phi = B A \cos \theta \] \[ \Phi = 4 \times (2.5 \times 2) \times \cos 60^\circ \] \[ \Phi = 4 \times 5 \times \frac{1}{2} = 10 \] Since the loop is removed completely, \[ \Delta \Phi = 10 - 0 = 10 \] Step 3: {Finding Average emf}
\[ \varepsilon = -\frac{10}{10} = -1 { V} \] Since the direction of emf is not specified, we take magnitude: \[ \varepsilon = 1 { V} \] Thus, the correct answer is \( +1 \) V.

Answer 2

Step 1: Use Faraday’s law of electromagnetic induction
The average emf induced in a loop is given by:
\[ \text{emf} = -\frac{\Delta \Phi_B}{\Delta t} \]
Where:
- \( \Phi_B = B A \cos\theta \) is the magnetic flux
- \( B \) = magnetic field (4 T)
- \( A \) = area of the loop
- \( \theta \) = angle between the normal to the loop and magnetic field (60°)
- \( \Delta t = 10 \) s

Step 2: Calculate area of the loop
Length = 2.5 m, Width = 2 m
\[ A = 2.5 \times 2 = 5 \, \text{m}^2 \]

Step 3: Calculate initial and final magnetic flux
Initial flux:
\[ \Phi_{initial} = B A \cos\theta = 4 \times 5 \times \cos(60^\circ) = 20 \times \frac{1}{2} = 10 \, \text{Wb} \]
Final flux = 0 (loop is completely removed from the field)
\[ \Delta \Phi_B = 0 - 10 = -10 \, \text{Wb} \]

Step 4: Calculate the average emf
\[ \text{emf} = -\frac{\Delta \Phi_B}{\Delta t} = -\left(\frac{-10}{10}\right) = +1 \, \text{V} \]

Final Answer:
The average emf induced in the loop is:
+1 V