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 (μ0 = permeability of vacuum, χ = magnetic susceptibility)

Question

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 (μ0 = permeability of vacuum, χ = magnetic susceptibility) (A) μ0 nI (1 − χ) (B) μ0 nIχ (C) μ0 nI2 (1 + χ) (D) μ0 nI (1 + χ)

Tardigrade Admin · Accepted Answer

Magnetic field due to solenoid =μ × n × I where μ= permeability n = turn density I = current Also μ=μ0(1+ chi) where chi= magnetic susceptibility B text solenoid =μ0 n I (1+ chi)

Q. 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 ( $μ_{0} =$ permeability of vacuum, $χ =$ magnetic susceptibility)

$μ_{0} n I (1 - χ)$

$μ_{0} n I χ$

$μ_{0} n I^{2} (1 + χ)$

$μ_{0} n I (1 + χ)$

Magnetic field due to solenoid $= μ \times n \times I$
where $μ =$ permeability
$n =$ turn density
$I =$ current
Also $μ = μ_{0} (1 + χ)$
where $χ =$ magnetic susceptibility
$B_{solenoid} = μ_{0} n I (1 + χ)$

Q. 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 (μ0​= permeability of vacuum, χ= magnetic susceptibility)

μ0​nI(1−χ)

μ0​nIχ

μ0​nI2(1+χ)

μ0​nI(1+χ)