Calculate the energy associated with the magnetic field of a \[200\]-turn solenoid in which a current of \[1.75{\text{ A}}\] produces a flux of \[3.70 \times {10^{ - 4}}{\text{Wb}}\] in each turn
\[(A)0.0324{\text{ J}}\]
\[(B)\;0.0648{\text{ J}}\]
\[(C)0.648{\text{ J}}\]
\[(D)0.324{\text{ J}}\]

Hint: This question can be solved by understanding the relation between flux and inductance. First, obtain the value of inductance from the flux formula and then substitute this value in the magnetic energy formula. The magnetic energy formula is a relation between the number of turns, the inductance, and the square of the current. Using this relation we can determine the energy associated with the magnetic field of the solenoid.

Formula used:
$\phi = LI$
${\text{Magnetic energy = }}\dfrac{1}{2}{\text{NL}}{{\text{I}}^{\text{2}}}$
Where $L$ stands for inductance, $I$ stands for current, $N$ stands for the number of turns


Complete step by step answer:
Let us mention the given values;
\[{\text{The number of turns}} = 200\]
\[{\text{Current, I}} = 1.75{\text{ A}}\]
\[{\text{flux, }}\phi {\text{ = }}3.70 \times {10^{ - 4}}{\text{Wb}}\]
To solve this question let us first determine the value of inductance.
We can do this by using the relation between flux, inductance, and current. Then by substituting the given values for flux and current, we can determine the value of inductance, L.
$\phi = LI$
\[3.70 \times {10^{ - 4}}{\text{ = L}} \times {\text{1}}{\text{.75 }}\]
Therefore by rearranging we get, \[{\text{L}} = \dfrac{{3.70 \times {{10}^{ - 4}}{\text{ }}}}{{1.75}}\]
Solving the above expression we get, Inductance, ${\text{L}} = 2.1143 \times {10^{ - 4}}{\text{Wb}}{{\text{A}}^{ - 1}}$
The next step is to determine the magnetic energy. For this use the magnetic energy formula which is a relation between the magnetic energy, number of turns, inductance, and current.
Using the expression, ${\text{magnetic energy = }}\dfrac{1}{2}{\text{NL}}{{\text{I}}^{\text{2}}}$and by substituting the given and obtained values of the number of turns, current, and inductance, we can determine the value for magnetic energy.
${\text{magnetic energy = }}\dfrac{1}{2}{\text{NL}}{{\text{I}}^{\text{2}}}$
${\text{magnetic energy = }}\dfrac{1}{2} \times {\text{200}} \times {\text{2}}{\text{.1143}} \times {\text{1}}{{\text{0}}^{ - 4}} \times {1.75^{\text{2}}}$
Simplifying the expression we get, ${\text{magnetic energy = 0}}{\text{.0648 J}}$
We have obtained magnetic energy as${\text{0}}{\text{.0648 J}}$.
Therefore option $(B)$ is the correct option.

Note: The number of lines that pass through a closed surface is defined as magnetic flux. The tendency to oppose a change in electric current by an electric conductor is referred to as inductance. Since the coil’s inductance is a result of the magnetic flux around it, the larger the value of magnetic flux for a given current greater will be the value of inductance. The value of inductance can also increase with respect to the increase in the number of turns.