Consider the following statements about the magnetic field of current loops:
(A) For a tightly wound coil of \(N\) turns, the magnetic field at the centre is \(N\) times the field of a single turn.
(B) The magnetic field lines of a current loop start on one face of the loop and end on the other face.
(C) A long wire is bent into a semicircular arc, with its straight parts along the line of the diameter. These straight parts add nothing to the field at the centre of the arc.
(D) At the centre of a semicircular arc carrying a current, the fields due to different elements of the arc point in different directions.
Which of the above statements are incorrect? Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (D) only
The field of a loop or an arc is the sum of the fields \( d\vec{B} \) of its small current elements. Each \( d\vec{B} \) is given by the Biot-Savart law.
(A) For a tightly wound coil of \(N\) turns, the magnetic field at the centre is \(N\) times the field of a single turn. — True. Each turn has nearly the same radius, so each adds \(\mu_0I/2R\). The total is \(B=\mu_0NI/2R\).
(B) The magnetic field lines of a current loop start on one face of the loop and end on the other face. — False. Magnetic field lines have no starting or ending point. The field lines of a current loop form closed loops.
(C) A long wire is bent into a semicircular arc, with its straight parts along the line of the diameter. These straight parts add nothing to the field at the centre of the arc. — True. For each element of these straight parts, \(d\vec{l}\) and \(\vec{r}\) are parallel. So \(d\vec{l}\times\vec{r}=0\) and they give no field at the centre.
(D) At the centre of a semicircular arc carrying a current, the fields due to different elements of the arc point in different directions. — False. For every element of the arc, \(d\vec{l}\times\vec{r}\) points the same way, normal to the plane of the arc. So all the contributions add up.
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,”
\( B=\frac{\mu_0 NI}{2R} \)
Magnetic Field on the Axis of a Circular Current Loop – “… The magnetic field lines due to a circular wire form closed loops …”
Example 4.5 – “(a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|.”
Example 4.5 – “(b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. …”