A small plane loop carrying a steady current has a magnetic moment of \(0.5\ \mathrm{A\,m^{2}}\). The magnitude of the magnetic field at a point in the plane of the loop, 50 cm from its centre, is (the distance is much larger than the size of the loop; take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(4\times10^{-7}\ \mathrm{T}\)
Far from a current loop, its field is the same as that of a magnetic dipole of moment \( m \). For a point in the plane of the loop, at a distance \( x \) from its centre, the field is \( B=\mu_0m/4\pi x^{3} \).
Given:
\( m=0.5\ \mathrm{A\,m^{2}} \)
\( x=50\ \mathrm{cm}=0.5\ \mathrm{m} \)
\( \begin{aligned} B &= \frac{\mu_0}{4\pi}\,\frac{m}{x^{3}} \\ &= 10^{-7}\times\frac{0.5}{(0.5)^{3}} \\ &= 10^{-7}\times\frac{0.5}{0.125} \\ &= 4\times10^{-7}\ \mathrm{T} \end{aligned} \)
Therefore, the field at this point in the plane of the loop is \( 4\times10^{-7}\ \mathrm{T} \). On the axis, at the same distance, it would be twice as large.
NCERT (2026-27):
Circular current loop as a magnetic dipole – “… we obtain the result for B for a point in the plane of the loop at a distance x from the centre. For x >>R,”
\( \displaystyle B\simeq\frac{\mu_0}{4\pi}\,\frac{m}{x^{3}} \)\( \qquad [4.25(b)] \)