Question

A magnetic flux of $8\times {{10}^{-4}}$ weber is linked with each turn of a 200- turn coil when there is an electric current of 4 A in it. Calculate the self-inductance of the coil.

Answer 1

Hint: To solve this problem we will be discussing about self-inductance and by equating induced voltage from faraday’s law of electromagnetic induction and in terms of change of current with respect to time self-inductance of coil will be calculated.

Formula used:
Faraday’s Law, induced voltage $V=-\dfrac{Nd\phi }{dt}$
Induced voltage in terms of inductance $V=-L\dfrac{di}{dt}$

Complete Step by step solution:
Self-inductance is defined as the property of current carrying coil which resists the change in current flowing through it which is caused by self-induced emf produced in it. Emf will try to resist the rise of current as well as fall of current from the current value and its direction will be opposite to applied voltage.

Self-inductance of a coil can be derived by Faraday’s law of electromagnetic induction,
$V=-\dfrac{Nd\phi }{dt}$ here, N(number of turns) and $\dfrac{d\phi }{dt}$(small change in flux with respect to time).

The induced voltage V which is derived by Faraday’s law can also be expressed in terms of self-inductance of coil with rate of change of current with respect to time,
$V=-L\dfrac{di}{dt}$, L (self-inductance) and $\dfrac{di}{dt}$(small change in current with respect to time)

Now by equating both induced voltage formulas we have,
$\Rightarrow \dfrac{Nd\phi }{dt}=-L\dfrac{di}{dt}$
$\Rightarrow L=\dfrac{Nd\phi }{di}$ this formula will be used for small change in flux, but for large change in flux and current we have,
$\Rightarrow L=\dfrac{N\phi }{i}$
Now, according to the question $\phi $=$8\times {{10}^{-4}}$, N=200 and i=4 A
So by putting the values in above formula we have,
$\Rightarrow L=\dfrac{200\times 8\times {{10}^{-4}}}{4}$
$\Rightarrow L=4\times {{10}^{-2}}H$(in henries)

$\therefore $ The self-inductance of the coil with 200 turns is $L=4\times {{10}^{-2}}H$.

Note:
Self-induced emf in the coil can also be called as back emf because it is opposing the change in current in a circuit, so, to change current some work needs to be done against self-induced emf which is termed as magnetic potential energy.