Question

1. Calculate the inductance of a solenoid containing $$300$$ turns if the length of the solenoid is $$25cm$$ and its cross-sectional area is $4\text{c}{\text{m}}^2$.
2. Calculate the self-induced emf in this solenoid if the current is decreasing at the rate of $$50A/s$$.

Answer 1

Hint: When the current or magnetic flux of the coil changes, an opposing induced electromotive force is generated. The phenomenon is known as Self Induction. Induction is a magnetic field that is proportional to the magnetic field's rate of change. For a conductor, this definition of induction holds. Induction is referred to as inductance as well.

Formula Used: The formula for finding out the inductance of a solenoid is given by
$L={\displaystyle \frac{{\mu }_0{N}^{2}S}{l}}$
Where
$$L$$ is in Henries
$${\mu }_{\circ }$$ is the Permeability of Free Space ($$4.\pi .\times {10}^{-7}$$)
$$N$$ is the number of turns per unit length in the solenoid
$$S$$ is the inner core area ($$\pi r^2$$) in $$m^2$$
$$l$$ is the length of the Coil in metres
 The formula for self-induced emf is given by $$e=-L{\displaystyle \frac{di}{dt}}$$
Where
$e$ is the induced voltage
$$L$$ is the inductance (in henries)
$${\displaystyle \frac{di}{dt}}$$ is the rate of change of current

Complete Step-by-Step Solution:
 The inductance of a solenoid is given by
$L={\displaystyle \frac{{\mu }_0{N}^{2}S}{l}}$
According to the question we have the following data provided
permeability of free space, $${\mu }_{\circ }=4\pi \times {10}^{-7}$$
number of turns, $$N=300$$
inner core area, $$S=4.00\times {10}^{-4}$$
length of the coil, $$l=25.0\times {10}^{-2}$$
Substituting the values, we have
$L={\displaystyle \frac{\left(4\pi \times {10}^{-7}\right){(300)}^2\left(4.00\times {10}^{-4}\right)}{\left(25.0\times {10}^{-2}\right)}}H$
$\Rightarrow L=1.81\times {10}^{-4}H$
The formula for self-induced emf is given by
$e=-L{\displaystyle \frac{di}{dt}}$
Here according to the question
the inductance, $$L=1.81\times {10}^{-4}$$
Rate of change of current, ${\displaystyle \frac{di}{dt}}=-50.0\text{A}/\text{s}$
Now we will substitute the given data in the formula to get our required answer
$E=-\left(1.81\times {10}^{-4}\right)(-50.0)$
$\Rightarrow E=9.5\times {10}^{-3}V$
Hence we get,
$\therefore E=9.05\text{mV}$

Note: Do not get confused between induced emf and self-induced emf. Both these terms sound very similar, but in reality, they are very much different from each other. This question was easily solved by simply putting given values into the formula of induced emf and self-induced emf.