A small ball with a charge of \(2\ \mu\mathrm{C}\) is whirled in a horizontal circle of radius 10 cm. It makes 50 revolutions per second. The magnetic moment of the circulating charge is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(3.1\times10^{-6}\ \mathrm{A\,m^{2}}\)
The moving charge acts like a current loop. The charge passes any point of the circle 50 times each second, so the current is \( I=qf \). The magnetic moment of a loop of area \( A \) is \( m=IA \).
\( \begin{aligned} I &= qf \\ &= 2\times10^{-6}\times50 \\ &= 1\times10^{-4}\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} A &= \pi r^{2} \\ &= 3.14\times(0.1)^{2} \\ &= 3.14\times10^{-2}\ \mathrm{m^{2}} \end{aligned} \)
\( \begin{aligned} m &= IA \\ &= 1\times10^{-4}\times3.14\times10^{-2} \\ &\approx 3.1\times10^{-6}\ \mathrm{A\,m^{2}} \end{aligned} \)
NCERT (2026-27):
Electric Current – “… The net amount of charge flowing across the area in the forward direction in the time interval t, then, is q = q+– q–. This is proportional to t for steady current”
\( \displaystyle I=\frac{q}{t} \)\( \qquad (3.1) \)
“is defined to be the current across the area in the forward direction. …”
Torque on a rectangular current loop in a uniform magnetic field – “… We define the magnetic moment of the current loop as,”
\( m=IA \)\( \qquad (4.22) \)