Question

A magnetic field of flux density 1.0 Wb m^-2 acts normal to a 80 turns coil of 0.01 m² area. If this coil is removed from the field in 0.2 second, the emf induced in it is

• A. 0.8 V
• B. 4 V
• C. 5 V
• D. 8 V

Answer 1

The Correct Option is B

Faraday's law of electromagnetic induction:

\( \text{emf} = -N \cdot \frac{d\Phi}{dt}\)

In this case, we are given:

• Flux density (B) = 1.0 Wb/m²
• Area of the coil (A) = 0.01 m²
• Number of turns (N) = 80
• Time taken (dt) = 0.2 seconds

The magnetic flux (Φ) passing through the coil is given by:

\( \Phi = B \cdot A\)

\( \Phi = 1.0 \, \text{Wb/m}^2 \times 0.01 \, \text{m}^2\)

\( \Phi = 0.01 \, \text{Wb}\)

Now, we can calculate the rate of change of magnetic flux \( \left(\frac{d\Phi}{dt}\right) \):

\( \frac{d\Phi}{dt} = \frac{\Phi_{\text{final}} - \Phi_{\text{initial}}}{dt}\)

Since the coil is removed from the field, the final magnetic flux is zero (\( \Phi_{\text{final}} = 0 \)), and the initial magnetic flux (\( \Phi_{\text{initial}} \)) is 0.01 Wb. Substituting these values:

\( \frac{d\Phi}{dt} = \frac{0 \, \text{Wb} - 0.01 \, \text{Wb}}{0.2 \, \text{s}}\)

\( \frac{d\Phi}{dt} = -0.05 \, \text{Wb/s}\)

Finally, substituting the values into the emf formula:

\( \text{emf} = -N \cdot \frac{d\Phi}{dt}\)

\(\text{emf} = -80 \times (-0.05 \, \text{Wb/s})\)

\(\text{emf} = 4 \, \text{V}\)

Therefore, the emf induced in the coil is 4 V. Option (B) is correct.

Answer 2

The induced EMF in a coil is given by Faraday's law of electromagnetic induction:

\(\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t}\)

Where:

• \(\mathcal{E}\) is the induced EMF
• \(N\) is the number of turns in the coil
• \(\Delta \Phi\) is the change in magnetic flux
• \(\Delta t\) is the time interval over which the change occurs

The magnetic flux (\(\Phi\)) is given by:

\(\Phi = B \cdot A = BA \cos \theta\)

Where:

• \(B\) is the magnetic flux density
• \(A\) is the area of the coil
• \(\theta\) is the angle between the magnetic field and the normal to the coil area

Given:

• \(B = 1.0 \, \text{Wb/m}^2\)
• \(N = 80\) turns
• \(A = 0.01 \, \text{m}^2\)
• \(\Delta t = 0.2 \, \text{s}\)
• The magnetic field is normal to the coil, so \(\theta = 0^\circ\) and \(\cos \theta = 1\)

Initially, the magnetic flux through the coil is:

\(\Phi_1 = B A = (1.0 \, \text{Wb/m}^2)(0.01 \, \text{m}^2) = 0.01 \, \text{Wb}\)

When the coil is removed from the field, the final magnetic flux is:

\(\Phi_2 = 0 \, \text{Wb}\)

The change in magnetic flux is:

\(\Delta \Phi = \Phi_2 - \Phi_1 = 0 - 0.01 = -0.01 \, \text{Wb}\)

Now, calculate the induced EMF:

\(\mathcal{E} = -N \frac{\Delta \Phi}{\Delta t} = -80 \frac{-0.01}{0.2} = \frac{80 \times 0.01}{0.2} = \frac{0.8}{0.2} = 4 \, \text{V}\)

Therefore, the induced EMF in the coil is 4 V.