A circular loop of one turn has a radius of 10 cm. Its magnetic moment is \(0.314\ \mathrm{A\,m^{2}}\). The magnitude of the magnetic field at the centre of the loop is (take \(\mu_0=4\pi\times10^{-7}\ \mathrm{T\,m\,A^{-1}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2\pi\times10^{-5}\ \mathrm{T}\)
The magnetic moment of a loop is \( m=IA \), where \( A=\pi R^{2} \) is the area of the loop. First find the current from \( m \). Then \( B_0=\mu_0I/2R \) gives the field at the centre.
Here \( R=10\ \mathrm{cm}=0.1\ \mathrm{m} \).
\( \begin{aligned} I &= \frac{m}{\pi R^{2}} \\ &= \frac{0.314}{3.14\times0.01} \\ &= 10\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} B_0 &= \frac{\mu_0 I}{2R} \\ &= \frac{4\pi\times10^{-7}\times10}{2\times0.1} \\ &= 2\pi\times10^{-5}\ \mathrm{T} \end{aligned} \)
NCERT (2026-27):
Magnetic Field on the Axis of a Circular Current Loop – “As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain,”
\( B_0=\frac{\mu_0 I}{2R} \)\( \qquad (4.12) \)
Torque on a rectangular current loop in a uniform magnetic field – “… We define the magnetic moment of the current loop as, m = I A …”