A circular loop of wire has a radius of 10 cm. A charge of 6 C flows through the loop in 1 minute at a steady rate. 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 3
The correct answer is Option (3) → \(6.3\times10^{-7}\ \mathrm{T}\)
First find the current. Current is the charge that flows per second, \(I=q/t\). Then the field at the centre of the loop is \(B_0=\mu_0I/2R\).
\(q=6\ \mathrm{C}\)
\(t=1\ \mathrm{min}=60\ \mathrm{s}\)
\(R=10\ \mathrm{cm}=0.1\ \mathrm{m}\)
\( \begin{aligned} I &= \frac{q}{t} \\ &= \frac{6}{60} \\ &= 0.1\ \mathrm{A} \end{aligned} \)
\( \begin{aligned} B_0 &= \frac{\mu_0 I}{2R} \\ &= \frac{4\pi\times10^{-7}\times0.1}{2\times0.1} \\ &= 2\pi\times10^{-7}\ \mathrm{T} \\ &\approx 6.3\times10^{-7}\ \mathrm{T} \end{aligned} \)
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,”
\( B_0=\frac{\mu_0 I}{2R}\,\hat{i} \)\( \qquad (4.12) \)