Magnetic Effects of Current — Oersted to Ampere

The Discovery That Changed Everything

In 1820, Hans Christian Oersted noticed something nobody expected: a compass needle deflected when placed near a current-carrying wire. This single observation — that electricity and magnetism are connected — launched the entire field of electromagnetism.

The key insight is this: a moving charge creates a magnetic field. When current flows through a wire, billions of electrons move together, and their collective motion generates a magnetic field that wraps around the wire like invisible rings.

This is why we study magnetic effects of current — not just because it appears in every board and competitive exam, but because it explains how motors, generators, MRI machines, and speakers actually work.

Key Terms and Definitions

Magnetic field (B): A region of space where a magnetic force acts on moving charges or magnetic poles. SI unit is Tesla (T). Also expressed in Gauss (1 T = 10,000 G).

Ampere’s circuital law: The line integral of magnetic field around any closed loop equals $\mu_0$ times the total current enclosed. More powerful than Biot-Savart for symmetric situations.

Biot-Savart Law: Gives the magnetic field $d\vec{B}$ due to a small current element $Id\vec{l}$ .

Solenoid: A long coil of wire that acts like a bar magnet when current flows — used in electromagnets and inductors.

Toroid: A solenoid bent into a doughnut shape; the magnetic field is confined entirely inside.

Ampere (unit): The SI unit of current, defined as the current that produces a force of $2 \times 10^{-7}$ N/m between two parallel wires 1 m apart.

Biot-Savart Law — The Fundamental Relation

d\vec{B} = \frac{\mu_0}{4\pi} \cdot \frac{I \, d\vec{l} \times \hat{r}}{r^2}

$\mu_0 = 4\pi \times 10^{-7}$ T·m/A (permeability of free space)
$I$ = current in amperes
$dl$ = length of current element
$r$ = distance from element to field point
Direction: right-hand rule on $d\vec{l} \times \hat{r}$

The Biot-Savart law tells us the magnetic field contribution from each tiny piece of current. We then integrate over the entire wire.

Magnetic Field Due to a Straight Wire

For a long straight wire carrying current $I$ , the field at perpendicular distance $r$ is:

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

The field lines are concentric circles around the wire. Use the right-hand thumb rule: point your thumb along the current direction, and your curled fingers show the field direction.

Magnetic Field at the Centre of a Circular Loop

B = \frac{\mu_0 I}{2R}

where $R$ is the radius. The field at the centre is perpendicular to the plane of the loop.

For $n$ turns: $B = \frac{n\mu_0 I}{2R}$ . Each turn contributes equally, so we simply multiply. This is the principle behind Helmholtz coils used in labs.

Ampere’s Circuital Law

\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}

$I_{enc}$ = total current passing through the surface bounded by the loop.

This is the magnetic equivalent of Gauss’s law for electricity. It’s only useful when the system has enough symmetry to pull $B$ out of the integral.

When to use Biot-Savart vs Ampere’s Law:

Biot-Savart: Any shape of current, but requires integration
Ampere’s Law: Only works for highly symmetric cases (infinite straight wire, solenoid, toroid) but gives results in seconds

Field Inside a Solenoid

For a solenoid with $n$ turns per unit length:

B = \mu_0 n I

The field inside is uniform and parallel to the axis. Outside is nearly zero for a long solenoid. This is why solenoids are used as electromagnets — all the magnetic energy is concentrated inside.

Field Inside a Toroid

B = \frac{\mu_0 N I}{2\pi r}

where $N$ is total turns and $r$ is the distance from the centre of the toroid. Outside the toroid, $B = 0$ .

Force on a Current-Carrying Conductor

When a conductor of length $L$ carrying current $I$ is placed in magnetic field $\vec{B}$ :

\vec{F} = I\vec{L} \times \vec{B}

F = BIL\sin\theta

Direction: Fleming’s left-hand rule (FBI rule)

Fleming’s left-hand rule: Point your left-hand fingers along $\vec{B}$ , curl them toward $I\vec{L}$ . Your thumb points along the force. In India, we also call this the “FBI rule” — Force, B-field, I-current.

Force between two parallel wires: Two parallel wires carrying currents $I_1$ and $I_2$ separated by distance $d$ :

\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

Same-direction currents attract; opposite-direction currents repel. This fact is used to define the ampere.

Moving Charges in Magnetic Fields

A charge $q$ moving with velocity $\vec{v}$ in field $\vec{B}$ experiences:

\vec{F} = q\vec{v} \times \vec{B}

This is the Lorentz force. Key properties:

The force is always perpendicular to velocity — so it changes direction but never does work
A charged particle in a uniform field moves in a circle
The radius of circular motion: $r = \frac{mv}{qB}$
The time period: $T = \frac{2\pi m}{qB}$ (independent of speed — this is the cyclotron principle)

JEE Main frequently asks about charged particles entering uniform fields at various angles. If the velocity is parallel to $\vec{B}$ , there is zero force. If it’s perpendicular, the particle moves in a circle. If it’s at an angle, the particle traces a helix.

The Moving Coil Galvanometer

A galvanometer works because a current-carrying coil in a magnetic field experiences a torque:

\tau = NBIA\cos\theta

At equilibrium, this equals the restoring torque from the spring: $\tau = k\phi$ where $\phi$ is the deflection angle.

\phi = \frac{NBIA}{k}

Deflection is proportional to current — making it a current-measuring device.

Converting to Ammeter: Add a low resistance (shunt) $S$ in parallel. Converting to Voltmeter: Add a high resistance $R$ in series.

Solved Examples

Example 1 — CBSE Level

A straight wire carries 5 A. Find the magnetic field at 10 cm from the wire.

B = \frac{\mu_0 I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.10}

B = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.10} = \frac{2 \times 10^{-6} \times 5}{0.10} = 1 \times 10^{-5} \text{ T}

\boxed{B = 10 \, \mu\text{T}}

Example 2 — JEE Main Level

A solenoid of length 0.5 m has 500 turns and carries 2 A. Find the field inside.

Number of turns per metre: $n = \frac{500}{0.5} = 1000$ turns/m

B = \mu_0 n I = 4\pi \times 10^{-7} \times 1000 \times 2

B = 8\pi \times 10^{-4} \approx 2.51 \times 10^{-3} \text{ T} \approx 2.51 \text{ mT}

Example 3 — JEE Advanced Level

Two long parallel wires 0.2 m apart carry currents of 10 A and 15 A in opposite directions. Find the field midway between them.

At the midpoint, fields from both wires add (since opposite currents produce fields in the same direction at the midpoint).

B_1 = \frac{\mu_0 \times 10}{2\pi \times 0.1} = \frac{4\pi \times 10^{-7} \times 10}{2\pi \times 0.1} = 2 \times 10^{-5} \text{ T}

B_2 = \frac{\mu_0 \times 15}{2\pi \times 0.1} = 3 \times 10^{-5} \text{ T}

B_{total} = B_1 + B_2 = 5 \times 10^{-5} \text{ T} = 50 \, \mu\text{T}

Common Mistakes to Avoid

Mistake 1: Using Biot-Savart formula for the solenoid instead of Ampere’s law. For a solenoid with many turns, $B = \mu_0 n I$ is the correct result. Don’t try to integrate Biot-Savart for each turn separately in an exam.

Mistake 2: Forgetting that force on a charge at rest in a magnetic field is zero. $F = qvB\sin\theta$ — if $v = 0$ , no force, no matter how strong the field.

Mistake 3: Confusing the direction of current in the right-hand rule. The conventional current flows from + to −, not the direction of electron flow.

Mistake 4: For circular motion in a magnetic field, students write $F = \frac{mv^2}{r}$ and then forget to set it equal to $qvB$ to find $r$ . Always start from $qvB = \frac{mv^2}{r}$ .

Mistake 5: In shunt problems, assuming the full current passes through the galvanometer. The shunt takes the extra current so only a fraction $I_g$ flows through the galvanometer.

Exam-Specific Tips

CBSE Class 10: Focus on the straight-wire formula, the rule for direction, and how a solenoid behaves. Numerical problems usually give current and distance and ask for $B$ .

CBSE Class 12: Biot-Savart, Ampere’s circuital law, force between wires, and the galvanometer conversion are all high-weightage. Each year 4-6 marks come from this chapter.

JEE Main 2024: A charged particle in a crossed electric and magnetic field appeared in Shift 1. Know the velocity selector condition: $qE = qvB \Rightarrow v = E/B$ .

NEET: Force on current-carrying conductors, torque on a current loop, and galvanometer conversion appear almost every year. Straightforward numericals are preferred.

Practice Questions

Q1. A circular loop of radius 0.1 m carries a current of 3 A. Find the magnetic field at the centre.

$B = \frac{\mu_0 I}{2R} = \frac{4\pi \times 10^{-7} \times 3}{2 \times 0.1} = 6\pi \times 10^{-6}$ T $\approx 18.85 \, \mu$ T

Q2. A solenoid has 2000 turns per metre and carries 1.5 A. Find B inside.

$B = \mu_0 n I = 4\pi \times 10^{-7} \times 2000 \times 1.5 = 12\pi \times 10^{-4}$ T $\approx 3.77$ mT

Q3. A proton moves at $2 \times 10^6$ m/s perpendicular to a field of 0.5 T. Find the radius of its circular path. (Mass of proton $= 1.67 \times 10^{-27}$ kg, charge $= 1.6 \times 10^{-19}$ C)

$r = \frac{mv}{qB} = \frac{1.67 \times 10^{-27} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 0.5} = \frac{3.34 \times 10^{-21}}{8 \times 10^{-20}} \approx 0.042$ m $= 4.2$ cm

Q4. Two parallel wires 4 cm apart carry 5 A each in the same direction. Find the force per unit length between them.

$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} = \frac{4\pi \times 10^{-7} \times 5 \times 5}{2\pi \times 0.04} = \frac{10^{-6} \times 25}{0.04} = 6.25 \times 10^{-5}$ N/m (attractive, same direction)

Q5. A galvanometer of resistance 50 Ω gives full deflection for 1 mA. How do we convert it to an ammeter reading 5 A?

Shunt resistance $S = \frac{I_g G}{I - I_g} = \frac{0.001 \times 50}{5 - 0.001} \approx \frac{0.05}{4.999} \approx 0.01$ Ω. Connect this shunt in parallel.

Q6. Find the magnetic field inside a toroid of 500 turns, mean radius 0.25 m, carrying 2 A.

$B = \frac{\mu_0 N I}{2\pi r} = \frac{4\pi \times 10^{-7} \times 500 \times 2}{2\pi \times 0.25} = \frac{4 \times 10^{-4}}{0.5} = 8 \times 10^{-4}$ T

Q7. A wire of length 0.5 m carrying 4 A is placed at 30° to a magnetic field of 0.3 T. Find the force on it.

$F = BIL\sin\theta = 0.3 \times 4 \times 0.5 \times \sin 30° = 0.6 \times 0.5 = 0.3$ N

Q8. An electron enters a uniform field of $10^{-3}$ T with velocity $5 \times 10^6$ m/s perpendicular to the field. Find the time period of circular motion.

$T = \frac{2\pi m}{qB} = \frac{2\pi \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 10^{-3}} = \frac{5.73 \times 10^{-30}}{1.6 \times 10^{-22}} \approx 3.58 \times 10^{-8}$ s $\approx 35.8$ ns

FAQs

Q: Why does a compass needle deflect near a current-carrying wire?

The current creates a magnetic field that exerts a torque on the compass needle (which is itself a small magnet). The needle aligns with the resultant of the Earth’s field and the wire’s field.

Q: What is the difference between magnetic field B and magnetic flux density?

In most contexts at this level, B is called both the magnetic field and the magnetic flux density — they’re the same quantity in free space. The distinction matters only in materials where we also define H (magnetic field intensity).

Q: Why is the force between parallel wires zero if they carry no current?

With no current, there are no moving charges, so no magnetic field is created. No field means no force. The wires might still exert gravitational and electrostatic forces, but those are negligible at normal scale.

Q: How does a cyclotron use the fact that time period is independent of velocity?

The alternating voltage that accelerates the particle is timed to the particle’s natural circular frequency. Since $T = \frac{2\pi m}{qB}$ doesn’t depend on speed, as the particle speeds up and spirals outward, it always takes the same time for each semicircle — so the alternating field stays in sync.

Q: Why is the field outside a toroid exactly zero?

By Ampere’s law applied to a loop outside the toroid: the currents going one way through the loop are exactly cancelled by those going the other way. Net enclosed current is zero, so $B = 0$ .