 The magnetic field (B) inside a long solenoid having ‘n’, turns per unit length and carrying current ‘I’ when iron core is kept in it is (${{\mu }_{0}}=$ permeability of vacuum, $\chi =$ magnetic susceptibility)A.${{\mu }_{0}}nI\left( 1-\chi \right)$B.${{\mu }_{0}}nI\chi$C.${{\mu }_{0}}n{{I}^{2}}\left( 1+\chi \right)$D.${{\mu }_{0}}nI\left( 1+\chi \right)$ Verified
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Hint: For such question, we know that magnetic field due to solenoid is given by, $\mu \times n\times I$ and relation between magnetic field and magnetic susceptibility can be given as, $\mu ={{\mu }_{0}}\left( 1+\chi \right)$, using this two formulas of magnetic field and magnetic susceptibility we will find the answer.

Formula used:
$B=\mu \times n\times I$, $\mu ={{\mu }_{0}}\left( 1+\chi \right)$

In the question we are given that a long solenoid having n turns is kept in magnetic field so, first of all the equation of magnetic field of solenoid having n turns is given by,
$B=\mu \times n\times I$ …………………….(i)
Where, B is magnetic field, n is number of terms, I is current, and $\mu$ is magnetic permeability.
Now, magnetic permeability can be defined as the ratio of the magnetic induction to the magnetic intensity. In which magnetic induction is the magnetic field induced in a current carrying conductor and magnetic intensity is the intensity of magnetic field up to a certain distance.
Now, magnetic permeability can also be given by the formula,
$\mu ={{\mu }_{0}}\left( 1+\chi \right)$
Where, ${{\mu }_{0}}$is permeability of vacuum, $\chi$ is magnetic susceptibility. Magnetic susceptibility can be defined as, A measure of how much a material will become magnetized in an applied magnetic field.
Now, substituting the value of $\mu$ in expression (i) we will get,
${{B}_{\text{solenoid}}}={{\mu }_{0}}nI\left( 1+\chi \right)$

So, the correct answer is “Option D”.