The magnetic field at the centre of a current-carrying arc is B = (mu0 I theta) / (4 pi R), where theta is the angle the arc makes at the centre in radians. It is just the full-loop field mu0 I / 2R multiplied by the fraction of the circle you have (theta / 2pi). Memory hook: an arc is a "slice" of a loop, so it gives that same slice-fraction of the loop's field.
An arc of radius R subtending angle theta at centre O gives B = (mu0 I theta)/(4 pi R). Common cases (semicircle, quarter, full loop) are shown; radial lead wires add nothing because dl is parallel to r-hat.
Your doubts, answered
What is the formula for the magnetic field at the centre of an arc?
B = (mu0 I theta) / (4 pi R). Here theta is the angle (in radians) that the arc subtends at the centre and R is the radius. This is the same as taking the full-loop field mu0 I / 2R and scaling it by the fraction theta / 2pi, because an arc is only a part of a full circle.
How do I get the semicircle and quarter-circle results quickly?
Just plug in the angle. For a semicircle theta = pi, so B = mu0 I / 4R (half of the loop value). For a quarter circle theta = pi/2, so B = mu0 I / 8R (one-fourth of the loop value). Always think 'what fraction of a full circle is this arc?'
Why do the straight radial wires connecting to an arc add nothing at the centre?
The magnetic field element is dB proportional to I dl x r-hat. For a straight wire that points along the radius (straight towards or away from the centre), dl and the position vector r-hat are parallel, so their cross product is zero. So radial lead wires give zero field at the centre. Only the curved arc contributes.
Should theta be in degrees or radians?
Radians. The formula B = (mu0 I theta)/(4 pi R) needs theta in radians. If a question gives 60 degrees, convert first: 60 degrees = pi/3 rad. Using degrees directly is the most common numerical mistake here.
What happens when the current splits into two arcs of one loop?
The current divides inversely with resistance, and resistance of an arc is proportional to its length (its angle). So the longer arc carries less current. The key result: for each arc, the product (current x angle) is the same, so both arcs give equal magnitude field. But they carry current in opposite senses around the centre, so their fields oppose and often cancel to zero.
⚠️ The NEET trap ✗ For a 90 degree quarter arc, plugging theta = 90 into B = (mu0 I theta)/(4 pi R) and getting a huge wrong number. ✓ Convert to radians first: 90 degrees = pi/2. Then B = (mu0 I)(pi/2)/(4 pi R) = mu0 I / 8R. The angle in this formula is ALWAYS in radians. 🧠 The single most common arc mistake in NEET.
Real NEET questions
2023
A very long conducting wire is bent in a semi-circular shape from A to B. The magnetic field at point P (centre of the semicircle) for the steady current configuration is given by:
A · mu0 i / 4R pointed into the page
B · mu0 i / 4R pointed away from the page
C · (mu0 i / 4R)(1 + 2/pi) pointed away from the page ✓
D · (mu0 i / 4R)(1 + 2/pi) pointed into the page
Solution: Step 1 (arc part): A semicircle subtends theta = pi at its centre, so B_arc = (mu0 i)(pi)/(4 pi R) = mu0 i / 4R. Step 2 (straight parts): The two collinear straight segments are each semi-infinite wires with P at perpendicular distance R from the end. Each gives B = mu0 i / (4 pi R). Two of them: 2 x mu0 i/(4 pi R) = mu0 i/(2 pi R) = (mu0 i/4R)(2/pi). Step 3 (add): All point out of the page, so B = (mu0 i/4R) + (mu0 i/4R)(2/pi) = (mu0 i/4R)(1 + 2/pi), away from the page. Answer C.
2019
A straight conductor carrying current i splits into two parts (arcs) of a circular loop of radius R and rejoins. The total magnetic field at the centre P of the loop is:
A · Zero ✓
B · 3 mu0 i / 32R, outward
C · 3 mu0 i / 32R, inward
D · mu0 i / 2R, inward
Solution: Step 1: The current splits into two arcs (angles theta1 and theta2) that are in parallel. Current divides inversely with resistance, and arc resistance is proportional to arc length (angle), so i1 theta1 = i2 theta2. Step 2: Field of an arc at centre is B = (mu0 i_arc theta)/(4 pi R). Since i1 theta1 = i2 theta2, both arcs give EQUAL magnitude field. Step 3: The two arcs carry current in opposite rotational senses about P, so their fields point opposite ways (one into, one out of page). Net field = 0. Answer A.
2026
A current I0 flows through a metallic circular loop of radius r. The resistance of segment ABC is half that of ADC. The magnitude of the magnetic field at the centre O of the loop is:
A · mu0 I0 / 12r ✓
B · mu0 I0 / 4r
C · mu0 I0 / 2r
D · mu0 I0 / 2 pi r
Solution: Step 1: Current divides inversely with resistance. With R_ABC = (1/2) R_ADC and I1 R_ABC = I2 R_ADC, we get I1 = 2 I2. Step 2: With I1 + I2 = I0, this gives I2 = I0/3 and I1 = 2 I0/3. Step 3: The two arcs carry current in opposite senses about O, so their central fields subtract: B = (mu0 I1)/(4r) - (mu0 I2)/(4r) = (mu0/4r)(2I0/3 - I0/3) = mu0 I0 / 12r. Answer A.
Solved Moving Charges And Magnetism NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the magnetic field at the centre of a semicircular loop?
B = mu0 I / 4R, which is exactly half the field of a full loop (mu0 I / 2R) because a semicircle is half a circle.
What is the field at the centre of a quarter circle arc?
B = mu0 I / 8R, one-fourth of the full loop field, since a quarter arc is one-fourth of a full circle (theta = pi/2).
Do the connecting straight wires ever contribute to the field at the centre?
Only if they are NOT pointing straight at the centre. Radial lead wires (pointing towards the centre) give zero. But offset straight segments, like the long collinear wires in the 2023 NEET question, do add a field.
How does current split between two arcs of a loop?
Inversely with resistance. Arc resistance is proportional to its length, so the shorter arc carries more current. Because current times angle is the same for both, their fields are equal in magnitude and often cancel.
Why is the arc field formula useful for NEET?
Almost every 'field at centre' question in Moving Charges and Magnetism is really a combination of arcs and straight wires. Mastering B = (mu0 I theta)/(4 pi R) lets you solve semicircle, quarter-circle and split-loop problems in seconds.
