A small coil has a magnetic moment of \(0.2\ \mathrm{A\,m^{2}}\). The radius of the coil is much smaller than 10 cm. The magnetic field at a point on its axis, 10 cm from its centre, is (take \(\mu_0/4\pi=10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(4\times10^{-5}\ \mathrm{T}\)
At a large distance on its axis, a small current loop behaves like a magnetic dipole. Its field there is \( B=\frac{\mu_0}{4\pi}\cdot\frac{2m}{x^{3}} \).
Here \( m=0.2\ \mathrm{A\,m^{2}} \) and \( x=10\ \mathrm{cm}=0.1\ \mathrm{m} \).
\( \begin{aligned} B &= \frac{\mu_0}{4\pi}\cdot\frac{2m}{x^{3}} \\ &= 10^{-7}\times\frac{2\times0.2}{(0.1)^{3}} \\ &= 4\times10^{-5}\ \mathrm{T} \end{aligned} \)
NCERT (2026-27):
Circular current loop as a magnetic dipole – “As earlier, we define the magnetic moment m to have a magnitude IA, m = I A. Hence,”
\( B\simeq\frac{\mu_0}{4\pi}\,\frac{2m}{x^{3}} \)\( \qquad [4.25(a)] \)