A solenoid has 800 turns per metre and carries a current of 0.5 A. Its core is made of a material of relative permeability 301. The magnetising current \(I_m\) for this solenoid is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(150\ \mathrm{A}\)
The magnetising current \( I_m \) is an extra current passed through the windings when the core is taken out. It makes the field the same as it was with the core. With the core the field is \( B=\mu_r\mu_0nI \). Without it, the current \( (I+I_m) \) gives \( B=\mu_0n(I+I_m) \).
Current, \( I=0.5\ \mathrm{A} \)
Relative permeability, \( \mu_r=301 \)
\( \begin{aligned} \mu_0n(I+I_m) &= \mu_r\mu_0nI \\ I+I_m &= \mu_rI \end{aligned} \)
\( \begin{aligned} I_m &= (\mu_r-1)I \\ &= (301-1)\times0.5 \\ &= 150\ \mathrm{A} \end{aligned} \)
Therefore, \( I_m \) does not depend on the number of turns per metre.
NCERT (2026-27):
Example 5.5 – “(d) The magnetising current IM is the additional current that needs to be passed through the windings of the solenoid in the absence of the core which would give a B value as in the presence of the core. …”
Magnetisation and Magnetic Intensity – “Consider a long solenoid of n turns per unit length and carrying a current I. The magnetic field in the interior of the solenoid was shown to be given by”
\( B_0=\mu_0nI \)\( \qquad (5.8) \)
Example 5.5 – “(b) The magnetic field B is given by”
\( B=\mu_r\mu_0H \) …