Question

A vertical solenoid has $200$ turns in a length of $0.4m$ and carries a current of $3A$ in anticlockwise direction. The flux density in the middle is about?

Answer 1

In order to solve this question, we should know that flux density here means to find magnetic field and here we will use the general formula of magnetic field due to solenoid at middle point and by using the value of given parameters we will solve for magnetic field due to solenoid at midpoint.

Formula used:
If N is the total number of turns in a solenoid, I is the current flowing through the solenoid and L is the total length of the solenoid then the magnetic field density B at middle part of the solenoid is calculated as $B = {\mu _o}\dfrac{{NI}}{L}$ where ${\mu _o} = 1.26 \times {10^{ - 6}}N{A^{ - 2}}$ known as the permeability of free space.

Complete step by step solution:
According to the question, we have given that
$N = 200$ number of turns in a solenoid.
$L = 0.4m$ total length of the solenoid
$I = 3A$ current flowing across the solenoid
${\mu _o} = 1.26 \times {10^{ - 6}}N{A^{ - 2}}$ so using the formula of magnetic field at middle of solenoid as $B = {\mu _o}\dfrac{{NI}}{L}$ and on putting the values of all parameters we get,
$B = 1.26 \times {10^{ - 6}} \times \dfrac{{200 \times 3}}{{0.4}}$
solving for B we get,
$B = 1.26 \times {10^{ - 6}} \times 1500$
$\Rightarrow B = 1.89 \times {10^{ - 3}}T$
Hence, the magnetic field flux density at the middle part of the given solenoid has a magnitude of $1.89 \times {10^{ - 3}}T$.

Note:
It should be remembered that, the Standard unit of magnetic field is Tesla while if we write it in terms of fundamental units, the unit of magnetic field will be $N{L^{ - 1}}{A^{ - 1}}$ and remember, the flux density due to solenoid is not uniform throughout the solenoid the magnitude of flux density at the end points of solenoid on each side has a magnitude of ${B_{end}} = \dfrac{{{B_{middle}}}}{2}$ which shows it’s half at the ends point as compared to middle part of the solenoid.