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: Apply Faraday's Law.
\[{Average emf}, \varepsilon = -\frac{\Delta \Phi}{\Delta t}\]Step 2: Calculate the Magnetic Flux.
\[\Phi = B A \cos \theta\]\[\Phi = 4 \times (2.5 \times 2) \times \cos 60^\circ\]\[\Phi = 4 \times 5 \times \frac{1}{2} = 10\]As the loop is completely removed:\[\Delta \Phi = 10 - 0 = 10\]Step 3: Determine the Average emf.
\[\varepsilon = -\frac{10}{10} = -1 { V}\]As the direction of emf is not specified, we consider its magnitude:\[\varepsilon = 1 { V}\]Therefore, the correct answer is \( +1 \) V.