A tightly wound circular coil has 50 turns and a radius of 10 cm. The current needed to produce a magnetic field of \(\pi\times10^{-4}\ \mathrm{T}\) at its centre 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) → 1 A
One circular turn gives a field \(\mu_0I/2R\) at its centre. In a tightly wound coil every turn has nearly the same radius, so the fields of the \(N\) turns add up to \(B=\mu_0NI/2R\).
\( \begin{aligned} I &= \frac{2RB}{\mu_0 N} \\ &= \frac{2\times0.1\times\pi\times10^{-4}}{4\pi\times10^{-7}\times50} \\ &= \frac{2\pi\times10^{-5}}{2\pi\times10^{-5}} \\ &= 1\ \mathrm{A} \end{aligned} \)
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,”
\( B_0=\frac{\mu_0 I}{2R}\,\hat{i} \)\( \qquad (4.12) \)
Example 4.6 – “Since the coil is tightly wound, we may take each circular element to have the same radius … The magnitude of the magnetic field is,”
\( B=\frac{\mu_0 NI}{2R} \)