JEE Mains Physics · Moving Charges and Magnetism

Field of Straight Wires and Arcs

A straight piece of wire gives (μ₀I/4πd)(sin α₁ + sin α₂) at a point a distance d from it, and an arc gives μ₀Iθ/4πR at its centre; every shape of wire is a sum of these pieces.

All Moving Charges and Magnetism notes JEE Mains Physics strategy

Why this matters

Twenty-six PYQs, fifteen of them multiple choice, and four from 2026. Six add the fields of two long parallel wires. Six find the field at the centre of a triangle, square or hexagon of wire, and two use the Biot–Savart law for a small element or for one moving charge. Twelve bend a wire into arcs and straight pieces, and every one of those comes with a figure: each piece is one standard result, and the work is adding them with the right signs.

Concept 1 of 3: Field of a long straight wire and of two parallel wires

The field lines of a long straight current are circles round the wire. Grip the wire with the right hand, thumb along the current, and the fingers curl the way the field points. The field falls off as one over the distance. With two wires, work out each field and its direction at the point, then add them as vectors.

Definition

Long straight wire: $B = \dfrac{\mu_0 I}{2\pi d}$ , with $\dfrac{\mu_0}{2\pi} = 2 \times 10^{-7}\ \text{T m/A}$ .
Direction: right-hand grip rule. At a point to one side of the wire the field is perpendicular to the line joining the point to the wire.
Two parallel wires, point between them: opposite currents give fields in the SAME direction, so they add; like currents give opposite fields, so they subtract (zero at the midpoint if the currents are equal).
Point outside the pair: the rule reverses. Opposite currents subtract, like currents add.
Point not on the line of the wires: the two fields are at an angle; if the lines from the point to the two wires are perpendicular, $B = \sqrt{B_1^{2} + B_2^{2}}$ .

Long straight wire

B = \frac{\mu_0 I}{2\pi d}

Worked example

Two long parallel wires 10 cm apart carry 20 A and 10 A in the same direction. Find the field (a) at the midpoint and (b) at a point in their plane 5 cm outside the 10 A wire. (

\mu_0/2\pi = 2 \times 10^{-7}

T m/A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 2 · Q13Moderate

Example 1 · Moving Charges and Magnetism · Field of Straight Wires and Arcs

A current of 30 A each flows in opposite directions in two conducting wires, placed parallel to each other at a distance of 8 cm . The magnetic field at the midpoint between the two wires is

\_\_\_\_

\mu T

. (

\frac{\mu_{0}}{4\pi}=10^{- 7}\text{ }N/A^{2}

)

Opposite currents add between the wires

Between two antiparallel currents both fields point the same way, so with equal currents the midpoint field is twice one wire's field. Subtracting them gives zero, which is the answer only for like currents.

Outside the pair the rule reverses

At a point beyond both wires, antiparallel currents give opposite fields and like currents give fields in the same direction. Draw each field's direction at the point before adding.

Fields at an angle add as vectors

When the point is not on the line of the wires, the two fields are not parallel. If the lines from the point to the wires are perpendicular, B = √(B₁² + B₂²), not B₁ + B₂.

Drill 5 more on field of a long straight wire and of two parallel wires

Concept 2 of 3: Field of a finite straight wire and of a polygon loop

The Biot–Savart law gives the field of each small piece of current. Added along a straight wire of finite length, it gives a formula that needs only the perpendicular distance d and the two angles at which the ends are seen. An infinite wire is the case where both angles are 90°. A polygon of wire is n equal straight sides, each giving the same field at the centre, all in the same direction.

Definition

Biot–Savart: $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^{2}}$ , direction along $d\vec l \times \hat r$ . A single charge q moving at v: $B = \dfrac{\mu_0}{4\pi}\dfrac{qv\sin\theta}{r^{2}}$ .
Finite straight wire: $B = \dfrac{\mu_0 I}{4\pi d}(\sin\alpha_1 + \sin\alpha_2)$ , with $\alpha_1, \alpha_2$ measured from the perpendicular dropped from the point to the wire.
Semi-infinite wire, point on the perpendicular through its end: $\alpha_1 = 90^{\circ}, \alpha_2 = 0$ , so $B = \dfrac{\mu_0 I}{4\pi d}$ , half the infinite-wire value.
A point on the line of the wire itself: $\theta = 0$ for every piece, so B = 0.
Regular polygon of n sides, side a: the centre is $d = \dfrac{a}{2\tan(\pi/n)}$ from each side, each side is seen at $\alpha = \pi/n$ on both ends, and the n fields add: $B = n\dfrac{\mu_0 I}{4\pi d}\,2\sin\dfrac{\pi}{n}$ .
Triangle: $d = \dfrac{a}{2\sqrt3}$ , $\alpha = 60^{\circ}$ . Square: $d = \dfrac{a}{2}$ , $\alpha = 45^{\circ}$ .

Finite straight wire

B = \frac{\mu_0 I}{4\pi d}\left(\sin\alpha_1 + \sin\alpha_2\right) \qquad dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^{2}}

Worked example

A square loop of side 20 cm carries 4 A. Find the magnetic field at its centre. (

\mu_0/4\pi = 10^{-7}

T m/A)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 23 Jan 2026 Shift 2 · Q4Moderate

Example 2 · Moving Charges and Magnetism · Field of Straight Wires and Arcs

The current passing through a conducting loop in the form of equilateral triangle of side

4\sqrt{3}\text{ }cm

is 2A. The magnetic field at its centroid is

\alpha \times 10^{- 5}\text{ }T

. The value of

\alpha

\_\_\_\_

. (Given :

\mu_{o} = 4\pi \times 10^{- 7}

SI units)

The angles are measured from the perpendicular

In B = (μ₀I/4πd)(sin α₁ + sin α₂), each α is the angle between the perpendicular from the point and the line to an end of the wire. Measuring it from the wire swaps sine for cosine.

A semi-infinite wire gives half, not the full value

At a point level with the end of a semi-infinite wire, the field is μ₀I/4πd. Using μ₀I/2πd, the infinite-wire result, doubles that piece's contribution.

The centre of a polygon is not at a distance a/2

For a triangle the centre is a/2√3 from each side; only for a square is it a/2. The general distance is a/(2 tan(π/n)).

Drill 7 more on field of a finite straight wire and of a polygon loop

Concept 3 of 3: Field at the centre of arcs and bent wires

A wire bent into arcs and straight pieces looks hard, but at the centre each piece is one standard result. List the pieces, write each one's size from the table, decide whether it points into or out of the page, and add with signs. A straight piece whose line passes through the point gives nothing at all.

Definition

An arc of angle $\theta$ (in radians) and radius R, at its centre: $B = \dfrac{\mu_0 I\theta}{4\pi R}$ . A full circle ( $\theta = 2\pi$ ) gives $\dfrac{\mu_0 I}{2R}$ .
Direction at the centre of an arc: curl the right-hand fingers along the current; the thumb gives the field (anticlockwise on the page: out of the page).
Straight pieces use the finite-wire formula; a piece pointing at the centre gives zero.
Two concentric arcs in one closed loop, on the same side, carry current round the centre in opposite senses, so their fields subtract.
When the fields of the pieces are not all along one line (pieces in different planes), add them as vectors.

Piece of wire	Field at the point	For I = 10 A at a distance or radius of 5 cm
Infinite straight wire, point at distance d	$\dfrac{\mu_0 I}{2\pi d}$	$40\ \mu\text{T}$
Semi-infinite wire, point on the perpendicular through its end	$\dfrac{\mu_0 I}{4\pi d}$	$20\ \mu\text{T}$ Half of the infinite wire, not the same.
Straight wire whose line passes through the point	Zero	Zero
Full circular loop, at its centre	$\dfrac{\mu_0 I}{2R}$	$40\pi\ \mu\text{T} \approx 126\ \mu\text{T}$
Three-quarter circle, at its centre	$\dfrac{3\mu_0 I}{8R}$	$30\pi\ \mu\text{T} \approx 94.2\ \mu\text{T}$
Semicircle, at its centre	$\dfrac{\mu_0 I}{4R}$	$20\pi\ \mu\text{T} \approx 62.8\ \mu\text{T}$
Quarter circle, at its centre	$\dfrac{\mu_0 I}{8R}$	$10\pi\ \mu\text{T} \approx 31.4\ \mu\text{T}$
Arc of angle θ in radians, at its centre	$\dfrac{\mu_0 I\theta}{4\pi R}$	$20\theta\ \mu\text{T}$

Every piece's field at the centre is perpendicular to the plane of the wire, so in a flat shape the pieces add or subtract along one line.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 8 · Q12Moderate

Example 3 · Moving Charges and Magnetism · Field of Straight Wires and Arcs

A hairpin like shape as shown in figure is made by bending a long current carrying wire. What is the magnitude of a magnetic field at point

P

which lies on the centre of the semicircle?

A straight piece aimed at the centre gives nothing

Radial straight pieces, and straight leads whose line passes through the centre, contribute zero. Leaving them out is correct; giving them μ₀I/4πd is not.

The angle of an arc must be in radians

B = μ₀Iθ/4πR uses θ in radians. A 90° arc is θ = π/2, giving μ₀I/8R; putting θ = 90 gives a field larger by a factor of about 57.

Check the sense of every piece before adding

Two arcs, or an arc and a straight lead, can give fields in opposite directions at the centre. Decide into or out of the page for each piece first; adding the sizes alone is the most common wrong option.

Drill 11 more on field at the centre of arcs and bent wires

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Field of a long straight wire and of two parallel wires
Long straight wire
$B = \frac{\mu_0 I}{2\pi d}$
Field of a finite straight wire and of a polygon loop
Finite straight wire
$B = \frac{\mu_0 I}{4\pi d}\left(\sin\alpha_1 + \sin\alpha_2\right) \qquad dB = \frac{\mu_0}{4\pi}\frac{I\,dl\sin\theta}{r^{2}}$

Reference tables (1)

Field at the centre of arcs and bent wires8 rows

Piece of wire	Field at the point	For I = 10 A at a distance or radius of 5 cm
Infinite straight wire, point at distance d	$\dfrac{\mu_0 I}{2\pi d}$	$40\ \mu\text{T}$
Semi-infinite wire, point on the perpendicular through its end	$\dfrac{\mu_0 I}{4\pi d}$	$20\ \mu\text{T}$ Half of the infinite wire, not the same.
Straight wire whose line passes through the point	Zero	Zero
Full circular loop, at its centre	$\dfrac{\mu_0 I}{2R}$	$40\pi\ \mu\text{T} \approx 126\ \mu\text{T}$
Three-quarter circle, at its centre	$\dfrac{3\mu_0 I}{8R}$	$30\pi\ \mu\text{T} \approx 94.2\ \mu\text{T}$
Semicircle, at its centre	$\dfrac{\mu_0 I}{4R}$	$20\pi\ \mu\text{T} \approx 62.8\ \mu\text{T}$
Quarter circle, at its centre	$\dfrac{\mu_0 I}{8R}$	$10\pi\ \mu\text{T} \approx 31.4\ \mu\text{T}$
Arc of angle θ in radians, at its centre	$\dfrac{\mu_0 I\theta}{4\pi R}$	$20\theta\ \mu\text{T}$

Every piece's field at the centre is perpendicular to the plane of the wire, so in a flat shape the pieces add or subtract along one line.

Watch out for (9)

Opposite currents add between the wires
→ Field of a long straight wire and of two parallel wires
Outside the pair the rule reverses
→ Field of a long straight wire and of two parallel wires
Fields at an angle add as vectors
→ Field of a long straight wire and of two parallel wires
The angles are measured from the perpendicular
→ Field of a finite straight wire and of a polygon loop
A semi-infinite wire gives half, not the full value
→ Field of a finite straight wire and of a polygon loop
The centre of a polygon is not at a distance a/2
→ Field of a finite straight wire and of a polygon loop
A straight piece aimed at the centre gives nothing
→ Field at the centre of arcs and bent wires
The angle of an arc must be in radians
→ Field at the centre of arcs and bent wires
Check the sense of every piece before adding
→ Field at the centre of arcs and bent wires

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