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.