Read the following passage and answer the questions.
A student studies the magnetic field produced by a steady current. She uses the Biot-Savart law. First she takes a short straight piece of wire, 1 cm long, placed at the origin. It carries a current of 5 A in the +x direction. She marks some points, each 10 cm from this piece. Some of these points lie on the x-, y- and z-axes. One point, Q, lies in the x-y plane, on a line that makes an angle of 30° with the +x-axis. She also moves a point P along the +y-axis to see how the field changes with distance. Next, she bends a long wire into a flat circular loop of radius \(R\) and passes a steady current \(I\) through it. She treats the loop as a large number of small current elements. She then studies the field at the centre of the loop and at points on its axis.
Consider the following statements about the circular loop:
(A) At the centre of the loop, the fields due to all the elements of the loop point in the same direction.
(B) At a point on the axis, the components of the fields perpendicular to the axis add up.
(C) The magnitude of the field at the centre of the loop is \(\mu_0 I/2\pi R\).
(D) The magnetic field lines of the loop form closed loops.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) and (D) only
Each small element of the loop gives a field \( d\vec B \) at right angles to the plane of \( d\vec l \) and \( \vec r \). The total field is the vector sum of all these fields.
(A) At the centre of the loop, the fields due to all the elements of the loop point in the same direction. — True. For every element, \( d\vec l \) and \( \vec r \) lie in the plane of the loop and are at right angles. So every \( d\vec B \) is normal to the plane, on the same side, and the fields add up.
(B) At a point on the axis, the components of the fields perpendicular to the axis add up. — False. The perpendicular component from one element is balanced by that from the element on the opposite side. Only the components along the axis add up.
(C) The magnitude of the field at the centre of the loop is \( \mu_0 I/2\pi R \). — False. Putting \( x=0 \) in the axial field gives \( B=\mu_0I/2R \). The value \( \mu_0I/2\pi R \) belongs to a long straight wire at distance \( R \).
(D) The magnetic field lines of the loop form closed loops. — True. The lines pass through the loop and return around the outside of the wire. They have no starting or ending point.
NCERT (2026-27):
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. …”
Magnetic Field on the Axis of a Circular Current Loop – “… When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result.”
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,”
\( \displaystyle B_0=\frac{\mu_0 I}{2R} \)\( \qquad (4.12) \)
Magnetic Field on the Axis of a Circular Current Loop – “The magnetic field lines due to a circular wire form closed loops …”