Question

If the current increases from 0 to 1A in 0.1 s in a coil of 5mH, then the magnitude of induced EMF is equal to:
A. 0.005V
B. 0.5V
C. 0.05V
D. 5V

Answer 1

Hint: We know that EMF generated is directly proportional to the negative rate of change of flux in that circuit \[(\phi = Li)\] . Again, we know that Faraday’s law states that the magnitude of the circulation of the electric field around a closed loop is equal to the rate of change of the magnetic flux through the area enclosed by the loop. So, we will write \[E = - \dfrac{{d\phi }}{{dt}}\]. Then we will put the value of \[\phi \] in the previous equation. This can be easily calculated by using the formula for EMF induced in an inductor. That’s how we will calculate this.

Complete Answer:
Here, we are given the rate of change of current and inductance of the coil. We know that the EMF generated is directly proportional to the negative rate of change of flux in that circuit. In the case of a current-carrying coil, the flux is given by:
\[\phi = Li\]
Where L is the self-inductance of the coil and i is the current flowing through the coil.
Now, substituting this in the EMF equation we get:
\[
E = - \dfrac{{d\phi }}{{dt}} \\
E = - \dfrac{{d(Li)}}{{dt}} \\
\] s
Here L is a constant and it can be taken out of the derivative.
\[E = - L\dfrac{{di}}{{dt}}\]
\[E = - 0.005\dfrac{1}{{0.1}}\]
E=0.05
The negative sign is to show the polarity of voltage only.
Therefore, the correct answer is option ‘C’.

Note: Value of inductance of a coil is given as \[\dfrac{{N\mu A}}{l}\] where N is the number of turns of the coil. A is the area of the coil’s face. L is the length of the coil and \[\mu\] is the permeability of the material placed inside the coil. If a soft iron core is placed inside the coil, it will greatly enhance the flux passing through it and will also increase the back EMF developed by the ends of the coil.