A closely wound solenoid has 400 turns and an area of cross-section \(5\times10^{-4}\ \mathrm{m^{2}}\). It carries a current of 2.5 A. It is placed in a uniform magnetic field of 0.4 T. Its magnetic moment makes an angle of \(120^\circ\) with the field. The magnetic potential energy of the solenoid is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(0.1\ \mathrm{J}\)
A closely wound solenoid carrying a current acts like a bar magnet of magnetic moment \( m=NIA \). In a uniform field its potential energy is \( U=-mB\cos\theta \), where \( \theta \) is the angle between \( \mathbf{m} \) and \( \mathbf{B} \).
Number of turns, \( N=400 \)
Area, \( A=5\times10^{-4}\ \mathrm{m^{2}} \)
Current, \( I=2.5\ \mathrm{A} \)
Magnetic field, \( B=0.4\ \mathrm{T} \)
Angle, \( \theta=120^\circ \)
\( \begin{aligned} m &= NIA \\ &= 400\times2.5\times5\times10^{-4} \\ &= 0.5\ \mathrm{A\,m^{2}} \end{aligned} \)
\( \begin{aligned} U &= -mB\cos\theta \\ &= -0.5\times0.4\times\cos120^\circ \\ &= -0.5\times0.4\times(-0.5) \\ &= 0.1\ \mathrm{J} \end{aligned} \)
The angle is more than \( 90^\circ \), so \( \cos\theta \) is negative. Hence, the potential energy of the solenoid is positive, \( 0.1\ \mathrm{J} \).
NCERT (2026-27):
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”
\( m=NIA \)\( \qquad (4.24) \)
The dipole in a uniform magnetic field – “The magnetic potential energy Um is given by”
\( U_m=\int\tau(\theta)d\theta=\int mB\sin\theta\,d\theta= \)\( -mB\cos\theta \)\( =-\mathbf{m}\cdot\mathbf{B} \qquad (5.3) \)