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

Answer: (B)

4 V

Answer 2

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:
$\Phi = B \cdot A$
Substituting the given values:
$\Phi = 1.0 \, \text{Wb/m}^2 \times 0.01 \, \text{m}^2$
$= 0.01 Wb$
Now, we can calculate the rate of change of magnetic flux $(\frac{dΦ}{dt})$:
$\frac{d\Phi}{dt} = \frac{\Phi_{\text{final}} - \Phi_{\text{initial}}}{dt}$
Since the coil is removed from the field, the final magnetic flux is zero (Φ_final = 0), and the initial magnetic flux (Φ_initial) is 0.01 Wb.
Substituting these values:
$\frac{d\Phi}{dt} = \frac{0 \, \text{Wb} - 0.01 \, \text{Wb}}{0.2 \, \text{s}}$

$= -0.01 Wb/0.2 s$
= -0.05 Wb/s
Finally, substituting the values into the emf formula:
$\text{emf} = -N \cdot \frac{d\Phi}{dt}$
= $-80 \times (-0.05 \, \text{Wb/s})$
= 4 V
Therefore, the emf induced in the coil is 4 V. Option (B) is correct.