Is Moving Charges and Magnetism part of the EAPCET Physics syllabus?

Yes. Moving Charges and Magnetism is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Physics questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.

What mistakes do students make in Moving Charges and Magnetism?

Using μ₀I/(2πr) for the centre of a circular loop — that is the long-wire result; the loop centre has μ₀I/(2R). Saying the magnetic force changes a particle’s kinetic energy — it is always perpendicular to the velocity, so it does no work. Applying the right-hand rule to an electron without reversing the direction — for a negative charge the force is opposite to v × B. Putting the shunt in series to make an ammeter — a small shunt goes in parallel with the galvanometer.

EAPCET Physics · Intermediate second year

Moving Charges and Magnetism

Moving Charges and Magnetism covers the force on moving charges and currents, the Biot–Savart and Ampère laws, and moving-coil galvanometers.

Subject: Physics
Taught in: Intermediate second year
Physics in the paper: 40 of 160 questions

Key formulas and results

$\vec{F} = q(\vec{v} \times \vec{B})$ ; radius of circular path $r = \frac{mv}{qB}$
Long straight wire: $B = \frac{\mu_{0}I}{2\pi r}$ ; centre of a circular loop: $B = \frac{\mu_{0}I}{2R}$
Long solenoid: $B = \mu_{0}nI$
Force per unit length between parallel wires: $\frac{\mu_{0}I_{1}I_{2}}{2\pi d}$
Cyclotron frequency: $f = \frac{qB}{2\pi m}$
Ammeter: galvanometer with a small shunt; voltmeter: galvanometer with a large series resistance
Shunt to raise a galvanometer’s range $n$ times ( $n = I/I_{g}$ ): $S = \frac{G}{n - 1}$
Series resistance to make a voltmeter of range $V$ : $R = \frac{V}{I_{g}} - G$
Torque on a coil: $\tau = NIAB\sin\theta$ ( $\theta$ between the normal and $\vec{B}$ ); magnetic moment $M = NIA$
Moving-coil galvanometer: $I = \frac{C\theta}{NAB}$ , where $C$ is the couple per unit twist
Field at the centre of a semicircular wire: $B = \frac{\mu_{0}I}{4R}$ ; finite straight wire: $B = \frac{\mu_{0}I(\sin\alpha + \sin\beta)}{4\pi d}$
Cyclotron: maximum kinetic energy $= \frac{q^{2}B^{2}r^{2}}{2m}$ ; radius $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}$
Force on a rectangular loop (sides parallel to a long wire at distances $a$ and $b$ ): $F = \frac{\mu_{0}I_{1}I_{2}l}{2\pi}\left(\frac{1}{a} - \frac{1}{b}\right)$
Finite straight wire: $B = \frac{\mu_{0}I}{4\pi d}(\sin\alpha + \sin\beta)$ ; semi-infinite wire, at a point opposite its end: $B = \frac{\mu_{0}I}{4\pi d}$
On the axis of a circular coil: $B = \dfrac{\mu_{0}NIR^{2}}{2(R^{2} + x^{2})^{3/2}}$
Centre of a circular arc subtending angle $\theta$ : $B = \frac{\theta}{2\pi}\cdot\frac{\mu_{0}I}{2R}$

What this chapter covers

Lorentz force
Motion of a charge in a magnetic field
Biot–Savart law
Ampère’s circuital law
Force between parallel currents
Galvanometer, ammeter and voltmeter

How it is usually asked

Magnetic field at a point
Charged particle in a field
Force between currents
Converting a galvanometer

Understanding Moving Charges and Magnetism

A charge moving through a magnetic field feels the force q(v × B), which is perpendicular to both v and B. Because this force is always perpendicular to the velocity, it changes the direction of motion but never the speed, so it does no work. A charge moving perpendicular to a uniform field therefore moves in a circle of radius mv/qB, with a period that does not depend on its speed — the principle behind the cyclotron.

Currents themselves produce magnetic fields. The Biot–Savart law gives the field of a small current element; Ampère’s circuital law gives the field quickly for symmetric cases such as a long wire or a solenoid. Parallel currents in the same direction attract. A moving-coil galvanometer becomes an ammeter with a small parallel shunt and a voltmeter with a large series resistance.

EAPCET questions usually ask for magnetic fields at a point, the radius and frequency of a charged particle’s circular motion, forces between current-carrying wires, and the shunt or series resistance needed to convert a galvanometer. If a charge enters at an angle to the field, split its velocity: the perpendicular part gives circular motion and the parallel part is unchanged, so the path becomes a helix.

Worked example

EAPCET-style question

A proton moves with a speed of $2 \times 10^{6}\ \text{m/s}$ perpendicular to a uniform magnetic field of 0.1 T. Taking the mass of the proton as $1.6 \times 10^{-27}\ \text{kg}$ and its charge as $1.6 \times 10^{-19}\ \text{C}$ , the radius of its circular path is

(1)0.2 m
(2)2 m
(3)0.02 m
(4)0.1 m

Solution

The magnetic force provides the centripetal force: $qvB = \frac{mv^{2}}{r}$ , so $r = \frac{mv}{qB}$ .
$r = \frac{1.6 \times 10^{-27} \times 2 \times 10^{6}}{1.6 \times 10^{-19} \times 0.1}$ .
$r = \frac{3.2 \times 10^{-21}}{1.6 \times 10^{-20}}$ .
$r = 0.2\ \text{m}$ .

Answer: 0.2 m

Shortcuts and results to remember

Neutral point of two parallel wires: same-direction currents give $x = \frac{r}{1 + I_{2}/I_{1}}$ from wire 1, between them; opposite currents give $x = \frac{r}{I_{2}/I_{1} - 1}$ , outside, beyond the wire with the smaller current.
A wire floats above a parallel wire when $\frac{\mu_{0}I_{1}I_{2}}{2\pi r}$ equals its weight per unit length $(m/l)g$ .
The force on a bent wire in a uniform field equals the force on the straight wire joining its ends.
Proton $(q, m)$ , deuteron $(q, 2m)$ and α-particle $(2q, 4m)$ : with equal kinetic energy their radii are $1 : \sqrt{2} : 1$ ; with equal speeds $1 : 2 : 2$ .
In a shunted galvanometer the currents divide as $I_{g}/I_{s} = S/G$ .
For a proton, deuteron and alpha particle, path radii are $1 : 2 : 2$ at the same speed, $1 : 1 : \tfrac{1}{2}$ at the same momentum, $1 : \sqrt{2} : 1$ at the same kinetic energy, and $1 : \sqrt{2} : \sqrt{2}$ after the same accelerating voltage.

Common mistakes to avoid

Using μ₀I/(2πr) for the centre of a circular loop — that is the long-wire result; the loop centre has μ₀I/(2R).
Saying the magnetic force changes a particle’s kinetic energy — it is always perpendicular to the velocity, so it does no work.
Applying the right-hand rule to an electron without reversing the direction — for a negative charge the force is opposite to v × B.
Putting the shunt in series to make an ammeter — a small shunt goes in parallel with the galvanometer.

How to prepare it

Use the right-hand rules for every direction question — guessing directions is where marks are lost.

Once the formulas feel familiar, test them the way the exam does: practise Moving Charges and Magnetism questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Moving Charges and Magnetism chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Moving Charges and Magnetism part of the EAPCET Physics syllabus?: Yes. Moving Charges and Magnetism is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Moving Charges and Magnetism for EAPCET?: Use the right-hand rules for every direction question — guessing directions is where marks are lost.
How many Physics questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Moving Charges and Magnetism?: Using μ₀I/(2πr) for the centre of a circular loop — that is the long-wire result; the loop centre has μ₀I/(2R). Saying the magnetic force changes a particle’s kinetic energy — it is always perpendicular to the velocity, so it does no work. Applying the right-hand rule to an electron without reversing the direction — for a negative charge the force is opposite to v × B. Putting the shunt in series to make an ammeter — a small shunt goes in parallel with the galvanometer.