Q. A bar magnet has coercivity $4 \times 10^{3}\ Am^{- 1}$. It is desired to demagnetise it by inserting it inside a solenoid $12\ cm$ long and having 60 turns. The current that should be sent through the solenoid is
A) $2\ A$
B) $4\ A$
C) $6\ A$
D) $8\ A$
Explanation
(d) The bar magnet has coercivity $4 \times 10^{3}\ Am^{- 1},\ i.e.,\ $it requires a magnetic intensity $H = 4 \times 10^{3}\ Am^{- 1}$ to get demagnetised. Let $i$ be the current carried by solenoid having $n$ number of turns per metre length, then by definition $H = ni$. Here $H = 4 \times 10^{3}amp\ turn\ metre^{- 1}$
$n = \frac{N}{l} = \frac{60}{0.12} = 500\ turn\ metre^{- 1}$
$\Rightarrow i = \frac{H}{n} = \frac{4 \times 10^{3}}{500} = 8.0\ A$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.