Question

A square loop of side 15 cm being moved towards right at a constant speed of 2 cm/s as shown in figure. The front edge enters the 50 cm wide magnetic field at t = 0. The value of induced emf in the loop at t = 10 s will be :

• A. 0.3 mV
• B. 4.5 mV
• C. zero
• D. 3 mV

Answer 1

The Correct Option is C

To determine the induced EMF in the loop at \(t = 10 \text{ s}\), we apply Faraday's law of electromagnetic induction. This law states that the induced EMF in a closed loop equals the negative rate of change of magnetic flux through it.

The formula for induced EMF \((\varepsilon)\) is:

\(\varepsilon = -\frac{{d\Phi}}{{dt}}\)

where \(\Phi\) is the magnetic flux.

Magnetic flux \((\Phi)\) through a coil is calculated as:

\(\Phi = B \times A\)

where \(B\) is the magnetic field strength and \(A\) is the area of the coil within the magnetic field.

Consider the square loop's position. The magnetic field is 50 cm wide, and the loop's side is 15 cm.

At \(t = 0\), the front edge of the loop begins entering the magnetic field.

By \(t = 10 \text{ s}\), the loop has traveled:

\(d = \text{velocity} \times \text{time} = 2 \, \text{cm/s} \times 10 \, \text{s} = 20 \, \text{cm}\)

Since the loop has moved 20 cm and its side is 15 cm, the entire loop is outside the magnetic field because \(20 \, \text{cm}\) is less than \(15 \, \text{cm} + 50 \, \text{cm}\). The loop has not fully entered the field, and it has not yet exited.

Therefore, the magnetic flux through the loop is zero at \(t=10\) s.

Consequently, the rate of change of magnetic flux \((\frac{d\Phi}{dt})\) is zero, resulting in an induced EMF of:

\(\varepsilon = 0 \, \text{V}\)

Thus, the induced EMF in the loop is zero.