A bar magnet of magnetic moment \(0.6\ \mathrm{A\,m^{2}}\) is to be replaced by an equivalent closely wound solenoid. The solenoid has 300 turns and an area of cross-section \(5\times10^{-4}\ \mathrm{m^{2}}\). The current that must flow in the solenoid is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(4\ \mathrm{A}\)
An equivalent solenoid has the same magnetic moment as the bar magnet. For a solenoid of \( N \) turns, each of area \( A \), carrying a current \( I \), the magnetic moment is \( m=NIA \).
Given:
\( m=0.6\ \mathrm{A\,m^{2}} \)
\( N=300 \)
\( A=5\times10^{-4}\ \mathrm{m^{2}} \)
\( \begin{aligned} I &= \frac{m}{NA} \\ &= \frac{0.6}{300\times5\times10^{-4}} \\ &= \frac{0.6}{0.15} \\ &= 4\ \mathrm{A} \end{aligned} \)
NCERT (2026-27):
Bar magnet as an equivalent solenoid – “… The magnetic moment of a bar magnet is thus equal to the magnetic moment of an equivalent solenoid that produces the same magnetic field.”
Torque on a rectangular current loop in a uniform magnetic field – “If the loop has N closely wound turns, the expression for torque, … still holds, with”
\( m=NIA \)\( \qquad (4.24) \)