Question

A square loop of side 2 cm enters a magnetic field with a constant speed of 2 cm s^-1 as shown. The front edge enters the field at t = 0s. Which of the following graph correctly depicts the induced emf in the loop?
( Take clockwise direction positive )

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is C

Given Information: 
Side of square loop, \( l = 2\,cm \)
Speed of loop entering the magnetic field, \( v = 2\,cm/s \)
Magnetic field, \( B \), is uniform and perpendicular to the plane of the loop.

Step-by-Step Explanation:

Step 1: Understanding the problem clearly:
When the loop enters the magnetic field, magnetic flux through the loop changes, inducing an emf according to Faraday’s law of electromagnetic induction.

Step 2: Determine how emf changes with time:
The induced emf is given by Faraday's law as:

\[ E = \frac{d\Phi}{dt} = B \cdot l \cdot v \]

Since \(B\), \(l\), and \(v\) are constant, the induced emf remains constant while the loop is entering or leaving the field. However, the emf changes direction based on whether the loop is entering or leaving the magnetic field.

Step 3: Calculate time intervals clearly:
The loop enters at speed \(2\,cm/s\), and loop side is \(2\,cm\). Thus, the time for the loop to completely enter the field is:

\[ t = \frac{\text{side}}{\text{speed}} = \frac{2\,cm}{2\,cm/s} = 1\,s \]

After fully entering, emf becomes zero since flux remains constant. Then, as the loop exits the field, emf is induced again (but with opposite sign), taking another \(1\,s\) to exit.

Step 4: Describe the induced emf clearly over time:

• From \(t=0\) to \(t=1\,s\), emf is constant positive (clockwise as per given sign convention).
• From \(t=1\,s\) to \(t=2\,s\), the emf is zero (flux is constant).
• From \(t=2\,s\) to \(t=3\,s\), emf is constant negative (opposite direction).

Thus, the correct graph should show a positive constant emf for the first interval, zero emf during the second interval, and negative constant emf in the third interval.

Conclusion:
Option (C) clearly matches this pattern.