Question

There is an iron core solenoid of turn density 10 tums/cm and volume 10 > m?. It carries a current of 0.5 A and the relative permeability of the iron core is u, = 1000. The magnetic moment of this solenoid is approximately (in A-m*)

Answer 1

Answer:

$we \: know \: that \: B \: = u_{o}H \: + u_{o}I$

$i \: = \frac{b - u_{o}H }{ u_{o} } = \frac{ u_{o}H - u_{o}h }{ u_{o} } = \binom{ u_{o} } { u_{o} - 1} h$

$I \: = ( u_{r} - 1)$

For a solenoid having n turns / length and current i,

H = ni

$I \: = ( u_{r} - 1)ni$

$= (1000 - 1)500 \times 0.5$

$= 2.5 \times {10}^{5} {Am}^{ - 1}$

∴Magnetic moment, M = IV

$= 2.5 \times {20}^{5} = {10}^{ - 4} = 25 {Am}^{2} .$

Answer 2

$\blue{\huge{\underline{\underline{Answer:-}}}}$

We know that, B=μ0H+μ0I

∴I=μ0B−μ0H=μ0μ0H−μ0H=(μ0μ0−1)H

I=(μr−1)H

For a solenoid having n turns/ length and current i,

H=ni

I=(μr−1)ni

=(1000−1)500×0.5

=2.5×105Am−1

∴ Magnetic moment, M=IV

=2.5×205=10−4=25 Am2.