A tightly wound circular coil has \(N\) turns and radius \(R\). It carries a current \(I\). In how many of the following cases does the magnetic field at its centre become double? (In each case, the quantities not named stay the same.)
(a) The current is doubled.
(b) The radius is halved.
(c) The number of turns is doubled.
(d) The radius is doubled.
(e) The current and the radius are both doubled.
(f) The current is made four times and the radius is doubled.
(g) The number of turns is halved and the current is doubled.
(h) The number of turns, the current and the radius are all doubled.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → Five
The field at the centre of a tightly wound coil is \( B=\mu_0NI/2R \). So \( B \) grows with \( N \) and \( I \), and falls as \( R \) grows.
The cases in which the field becomes double:
(a) \( I\to2I \) — \( 2B \)
(b) \( R\to R/2 \) — \( 2B \)
(c) \( N\to2N \) — \( 2B \)
(f) \( I\to4I,\ R\to2R \) — \( 4B/2=2B \)
(h) \( N\to2N,\ I\to2I,\ R\to2R \) — \( (2\times2/2)B=2B \)
The cases in which it does not:
(d) \( R\to2R \) — the field becomes \( B/2 \)
(e) \( I\to2I,\ R\to2R \) — the field stays \( B \)
(g) \( N\to N/2,\ I\to2I \) — the field stays \( B \)
Number of cases = 5
NCERT (2026-27):
Example 4.6 – “Since the coil is tightly wound, we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is,”
\( B=\frac{\mu_0 NI}{2R} \)