MHT-CET Physics · Magnetic Fields Due to Electric Current

Magnetic Field of Wires, Coils, Arcs, Solenoids and Toroids

Every piece of current makes a field by the Biot–Savart law; added up, a long straight wire gives μ₀I/(2πd) in circles around it, a circular coil gives μ₀NI/(2R) at its centre, an arc gives the same times its fraction of a circle, a half-infinite wire gives half the long-wire value, and a long solenoid or a toroid gives μ₀nI inside.

All Magnetic Fields Due to Electric Current notes MHT-CET Physics strategy

Why this matters

52 PYQs, 17 HARD — the largest page in the chapter and where most of its HARD questions sit. Twelve are straight wires — two parallel wires at a midpoint or at a point where their fields are perpendicular, inside and outside a thick wire, and where a wire cancels a loop; twenty are circular coils — the field at the centre and on the axis, coils in perpendicular planes, a wire rewound into more turns, and charges going round a circle; thirteen are arcs and bent wires read from a figure; seven are solenoids, toroids and displacement current. Four cards.

Concept 1 of 4: Field of Straight Wires

A long straight wire's field circles it with B = μ₀I/(2πd), falling as 1/d. With two wires, find each field's direction by the right-hand rule and add as vectors. Midway between them, like currents give OPPOSITE fields (they subtract, zero if equal) and opposite currents give fields in the SAME direction (they add). At a point where the lines to the two wires are perpendicular, the fields are perpendicular too, so add them by Pythagoras. Inside a thick wire the field rises linearly from zero, μ₀Ir/(2πR²); outside it falls as 1/r.

Definition

$B = \dfrac{\mu_0 I}{2\pi d}$ ; $\dfrac{x}{3}$ ⇒ 3B. 12 A giving $3 \times 10^{-5}$ T ⇒ d = 80 mm.
Midpoint, currents opposite: $B = \dfrac{\mu_0(I_1 + I_2)}{\pi d}$ ; same direction: $\dfrac{\mu_0(I_1 - I_2)}{\pi d}$ (zero if equal).
Perpendicular lines to wires d apart: each at $\dfrac{d}{\sqrt{2}}$ , $B = \dfrac{\mu_0}{2\pi r}\sqrt{I_1^2 + I_2^2}$ (8 A, 15 A, 7 cm ⇒ $68 \times 10^{-6}$ T).
Outside both, like currents 2r apart, P at r from the nearer: $\dfrac{\mu_0I}{2\pi}\left(\dfrac{1}{r} + \dfrac{1}{3r}\right) = \dfrac{2\mu_0I}{3\pi r}$ .
Thick wire radius R: inside $\dfrac{\mu_0Ir}{2\pi R^2}$ , outside $\dfrac{\mu_0I}{2\pi r}$ ; B(R/2) : B(3R) = 3 : 2.
Wire cancels a loop at its centre: $\dfrac{\mu_0I_c}{2R} = \dfrac{\mu_0I_w}{2\pi d}$ ⇒ $d = \dfrac{RI_w}{\pi I_c}$ .

Long straight wire

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

Worked example

Wires 10 cm apart carry 6 A and 4 A in opposite directions. Field midway?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 10th May Shift 1 · Q35Hard

Example 1 · Magnetic Fields Due to Electric Current · Magnetic Field of Current-Carrying Conductor

Two long parallel wires carrying currents 8 A and 15 A in opposite directions are placed at a distance of 7 cm from each other. A point P is equidistant from both the wires such that the lines joining the point to the wires are perpendicular to each other. The magnitude of magnetic field at point P is

(\sqrt{2} = 1.4)

(\mu_0 = 4\pi \times 10^{-7}\,\text{H/m})

Adding the fields of like currents at the midpoint

Midway between wires with currents in the SAME direction, their fields point opposite ways and subtract. Opposite currents add there.

Drill 11 more on field of straight wires

Concept 2 of 4: Field of a Circular Coil: Centre, Axis, and Rotating Charges

At the centre of a coil of N turns, B = μ₀NI/(2R), along the axis. On the axis at distance x it falls to μ₀NIR²/(2(R² + x²)^(3/2)) — at x = √3R, an eighth of the centre value. Rewinding a fixed wire into n turns shrinks the radius n times, so the centre field grows n². Two coils in perpendicular planes give perpendicular fields: add by Pythagoras. A charge q going round f times a second is a current qf, so it makes a field μ₀qf/(2r) at the centre — the same for a charged rotating ring.

Definition

Centre: $B = \dfrac{\mu_0NI}{2R}$ (n = 100 for 3.14 × 10⁻⁴ T at 0.4 A, 8 cm). Axis: $\dfrac{\mu_0NIR^2}{2(R^2 + x^2)^{3/2}}$ ; x = √3R ⇒ $\tfrac{1}{8}$ ; x = 2√2R ⇒ $\dfrac{\mu_0NI}{54R}$ .
Rewound into n turns: $B \times n^2$ (9 turns → 3 turns ⇒ B/9).
Perpendicular coils: $\sqrt{B_1^2 + B_2^2}$ (I and 2I ⇒ $\sqrt{5}\tfrac{\mu_0I}{2R}$ ; I and $\sqrt{8}I$ ⇒ $\tfrac{3\mu_0I}{2R}$ ).
Concentric coplanar coils: fields add or subtract with the current sense; to cancel, $\dfrac{I_A}{R_A} = \dfrac{I_B}{R_B}$ , opposite senses.
Rotating charge: $I = qf$ , $B = \dfrac{\mu_0qf}{2r}$ ; ring of charge at N r.p.s. ⇒ $q = \dfrac{2RB}{\mu_0N}$ ; orbiting electron $I = \dfrac{ev}{2\pi r}$ .
Same wire, two coils, equal centre fields, radii 2 : 1 ⇒ voltages 4 : 1.

Circular coil

B_{\text{centre}} = \frac{\mu_0 N I}{2R}, \qquad B_{\text{axis}} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}

Worked example

A 20-turn coil of radius 10 cm carries 0.5 A. Field at the centre, and on the axis 10√3 cm away?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q45Moderate

Example 2 · Magnetic Fields Due to Electric Current · Magnetic Field of Current-Carrying Conductor

A circular coil carrying current ' T ' has a radius '

r

' and '

n

' turns. The magnetic field along the axis of a coil at a distance '

2\sqrt{2}r

' from its centre is (

\mu_{0}=

permeability of free space,

n

is very small)

Adding fields of perpendicular coils directly

Coils in perpendicular planes have perpendicular axes, so their fields add as vectors: √(B₁² + B₂²), not B₁ + B₂.

Keeping the radius when a wire is rewound

The same wire wound into n turns has a radius n times smaller. The field μ₀NI/(2R) gains n from the turns AND n from the radius: n², not n.

Drill 19 more on field of a circular coil: centre, axis, and rotating charges

Concept 3 of 4: Arcs and Bent Wires

Break the wire into pieces and add their fields at the point. An arc subtending θ at its centre gives (μ₀I/4πR)·θ — a fraction θ/2π of a full coil. A straight piece whose LINE passes through the point gives nothing. A straight piece that ENDS at the foot of the perpendicular from the point and runs to infinity gives half a long wire, μ₀I/(4πr). Then decide each piece's direction (in or out of the page) by the right-hand rule: pieces going round the point the same way add, opposite ways subtract.

Definition

Arc of angle θ: $B = \dfrac{\mu_0I\theta}{4\pi R}$ ; semicircle $\dfrac{\mu_0I}{4R}$ , quarter $\dfrac{\mu_0I}{8R}$ , $\tfrac{3}{4}$ circle $\dfrac{3\mu_0I}{8R}$ , $\tfrac{\pi}{8}$ ⇒ $\dfrac{\mu_0I}{32R}$ .
Straight piece through the point: zero. Half-infinite piece ending at the foot: $\dfrac{\mu_0I}{4\pi r}$ . Long wire tangent to a loop: $\dfrac{\mu_0I}{2\pi r}$ .
Two semicircles R₁, R₂: $\dfrac{\mu_0I}{4}\left(\dfrac{1}{R_1} \pm \dfrac{1}{R_2}\right)$ — plus when both go round the centre the same way.
Loop in a long wire: $\dfrac{\mu_0I}{2\pi r}(\pi - 1)$ when the loop's field opposes the wire's, $(\pi + 1)$ when they agree.
Current element (Biot–Savart): $dB = \dfrac{\mu_0}{4\pi}\dfrac{I\,dl\sin\theta}{r^2}$ .

Arc and half-infinite wire

B_{\text{arc}} = \frac{\mu_0 I \theta}{4\pi R}, \qquad B_{\text{half-infinite}} = \frac{\mu_0 I}{4\pi r}

Worked example

A wire runs in from infinity along a line through O, goes round a semicircle of radius R about O, and leaves along the same line. Field at O?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 10th May Shift 2 · Q17Hard

Example 3 · Magnetic Fields Due to Electric Current · Magnetic Field of Current-Carrying Conductor

The magnitude of magnetic field at point 'O' in the following figure will be

Counting a straight piece that points at the centre

A straight wire whose line passes through the point gives zero field there, however long it is. Only pieces that pass BESIDE the point contribute.

Adding every piece without checking its direction

Two arcs carrying current round the centre in OPPOSITE senses give opposite fields: μ₀I/4 (1/R₂ − 1/R₁), not the sum. Fix each piece's in-or-out direction before adding.

Drill 12 more on arcs and bent wires

Concept 4 of 4: Solenoids, Toroids and Displacement Current

Inside a long solenoid the field is uniform, B = μ₀nI, with n the turns per metre; the magnetising field H = nI does not depend on what fills it. So B depends on n and I, not on the wire's thickness or the solenoid's radius. A toroid is a solenoid bent into a ring: B = μ₀NI/(2πr) = μ₀nI. Between capacitor plates a changing electric field acts as a current (the displacement current), spread over the plate area.

Definition

Solenoid: $B = \mu_0nI$ , $H = nI$ ; n × 3, I ÷ 4 ⇒ $\tfrac{3B}{4}$ . $H = \dfrac{N}{l}I$ (2.4 × 10³ A/m, 60 turns, 15 cm ⇒ 6 A).
Independent of the wire's radius and the solenoid's radius.
Toroid: $B = \dfrac{\mu_0NI}{2\pi r}$ (4000 turns, 5 A, 20 cm ⇒ $2 \times 10^{-2}$ T).
Displacement current through an area A/2 between plates charged by I: $\tfrac{I}{2}$ .

Solenoid and toroid

B = \mu_0 n I, \qquad B_{\text{toroid}} = \frac{\mu_0 N I}{2\pi r}

Worked example

A 0.5 m solenoid of 1000 turns carries 2 A. B inside (μ₀ = 4π × 10⁻⁷)?

Practice this conceptself-check · 2 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q32Easy

Example 4 · Magnetic Fields Due to Electric Current · Magnetic Field of Current-Carrying Conductor

The magnetic field intensity inside a current-carrying solenoid is

2.4 \times 10^3\,\text{Am}^{-1}

. If length and number of turns of a solenoid is 15 cm and 60 turns respectively. The current flowing in the solenoid is

Confusing B and H in a solenoid

H = nI (A/m) is set by the winding alone; B = μ₀nI (T) includes the medium. A question asking for H in an empty solenoid wants nI, not μ₀nI.

Drill 6 more on solenoids, toroids and displacement current

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 (4)

Field of Straight Wires
Long straight wire
$B = \frac{\mu_0 I}{2\pi d}$
Field of a Circular Coil: Centre, Axis, and Rotating Charges
Circular coil
$B_{\text{centre}} = \frac{\mu_0 N I}{2R}, \qquad B_{\text{axis}} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}}$
Arcs and Bent Wires
Arc and half-infinite wire
$B_{\text{arc}} = \frac{\mu_0 I \theta}{4\pi R}, \qquad B_{\text{half-infinite}} = \frac{\mu_0 I}{4\pi r}$
Solenoids, Toroids and Displacement Current
Solenoid and toroid
$B = \mu_0 n I, \qquad B_{\text{toroid}} = \frac{\mu_0 N I}{2\pi r}$

Watch out for (6)

Adding the fields of like currents at the midpoint
→ Field of Straight Wires
Adding fields of perpendicular coils directly
→ Field of a Circular Coil: Centre, Axis, and Rotating Charges
Keeping the radius when a wire is rewound
→ Field of a Circular Coil: Centre, Axis, and Rotating Charges
Counting a straight piece that points at the centre
→ Arcs and Bent Wires
Adding every piece without checking its direction
→ Arcs and Bent Wires
Confusing B and H in a solenoid
→ Solenoids, Toroids and Displacement Current

Test yourself on Magnetic Fields Due to Electric Current

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Related notes

Force on a Conductor — how these fields push on other wires →