Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page covers Moving Charges and Magnetism completely for CBSE Class 12 Physics.
Key Concepts
Moving Charges and Magnetism
Core concepts you must know
1
Biot-Savart Law
Formula
Determines the magnetic field $d\vec{B}$ at a point due to a small current element $Id\vec{l}$. The field is perpendicular to both the current element and the position vector. (1 Tesla = $10^4$ Gauss)
$$d \vec{B} = \frac{\mu_0}{4 \pi} \frac{I d\vec{l} \times \vec{r}}{r^3} \quad \text{or} \quad |d \vec{B}| = \frac{\mu_0}{4 \pi} \frac{I dl \sin \theta}{r^2}$$
2
Magnetic Field of Wires & Loops
Formula
Standard formulas for the magnetic field produced by a straight current-carrying conductor at distance $R$, and at the center of a circular loop of radius $a$ with $n$ turns.
$$\text{Straight Wire: } B = \frac{\mu_0 I}{2 \pi R} \quad | \quad \text{Loop Center: } B = \frac{\mu_0 n I}{2 a}$$
3
Field on the Axis of a Circular Loop
Formula
The magnetic field at a distance $x$ from the center of a current loop of radius $a$ along its axis. For far-away points ($x \gg a$), it behaves like a magnetic dipole ($B \propto 1/x^3$).
$$B = \frac{\mu_0 I}{4 \pi} \frac{2 \pi a^2}{(a^2 + x^2)^{3/2}} \xrightarrow{\text{n turns}} B = \frac{\mu_0 n I a^2}{2(a^2 + x^2)^{3/2}}$$
4
Ampere’s Circuital Law & Solenoids
Formula
The line integral of magnetic field over a closed loop equals $\mu_0$ times the total enclosed current. Used to find fields inside long straight solenoids and toroidal solenoids ($n$ is turns per unit length).
$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I \quad | \quad \text{Solenoid/Toroid: } B = \mu_0 n I$$
5
Lorentz Force
Formula
The total force experienced by a charged particle $q$ moving with velocity $\vec{v}$ in the presence of both electric field $\vec{E}$ and magnetic field $\vec{B}$.
$$\vec{F} = q\vec{E} + q(\vec{v} \times \vec{B})$$
6
Cyclotron Motion Parameters
Formula
When a charge enters a magnetic field perpendicularly, it moves in a circle. If it enters at an angle $\theta$, it moves in a helix. The time period is independent of the velocity!
$$r = \frac{mv}{qB} \quad | \quad T = \frac{2 \pi m}{Bq} \quad | \quad \nu = \frac{Bq}{2 \pi m} \quad | \quad E_{max} = \frac{B^2 q^2 r^2}{2m}$$
7
Force on Conductors & Parallel Wires
Formula
A current-carrying wire of length $\vec{l}$ experiences a force in a magnetic field. Two parallel wires carrying current exert force on each other (attractive if same direction, repulsive if opposite).
$$\vec{F} = I(\vec{l} \times \vec{B}) \quad | \quad \frac{F}{l} = f = \frac{\mu_0}{4 \pi} \frac{2 I_1 I_2}{r}$$
8
Magnetic Dipole Moment & Torque
Formula
A current loop acts as a magnetic dipole. In an external magnetic field, it experiences a torque trying to align it with the field. $\theta$ is the angle between area vector and $B$.
$$\vec{M} = I\vec{A} \quad | \quad \vec{\tau} = \vec{M} \times \vec{B} \quad | \quad \tau = nBIA \sin \theta$$
9
Moving Coil Galvanometer & Sensitivity
Formula
Works on the principle of torque on a current loop. The restoring torque $k\phi$ balances the deflecting torque. Sensitivities measure deflection per unit current/voltage.
$$I = \frac{k}{nBA} \phi = G\phi \quad | \quad I_s = \frac{\phi}{I} = \frac{nBA}{k} \quad | \quad V_s = \frac{\phi}{V} = \frac{nBA}{kR}$$
Concept Deep Dive
01
The Helical Path
What happens when velocity isn’t perfectly perpendicular?When a charged particle enters a magnetic field at an arbitrary angle $\theta$, its velocity resolves into two components. The perpendicular component ($v \sin \theta$) provides the centripetal force, making it move in a circle of radius $r = \frac{mv \sin \theta}{qB}$. The parallel component ($v \cos \theta$) remains completely unaffected by the magnetic field, dragging the particle forward at a constant speed. Combining circular motion with forward translation creates a helical path (like a spring). The distance moved forward in one full rotation is called the Pitch.
$$\text{Pitch} = (v \cos \theta) \times T = \frac{2\pi mv \cos \theta}{qB}$$
02
Galvanometer Conversion
Ammeter vs. VoltmeterA Moving Coil Galvanometer (MCG) is highly sensitive and cannot measure large currents or voltages directly without burning out.
To make an Ammeter: Connect a very small resistance (called a shunt, $S$) in parallel with the galvanometer. Most of the heavy current bypasses through the shunt, protecting the galvanometer. An ideal ammeter has zero resistance.
To make a Voltmeter: Connect a very large resistance ($R$) in series with the galvanometer. This ensures it draws negligible current from the circuit when connected in parallel across components. An ideal voltmeter has infinite resistance.
To make an Ammeter: Connect a very small resistance (called a shunt, $S$) in parallel with the galvanometer. Most of the heavy current bypasses through the shunt, protecting the galvanometer. An ideal ammeter has zero resistance.
To make a Voltmeter: Connect a very large resistance ($R$) in series with the galvanometer. This ensures it draws negligible current from the circuit when connected in parallel across components. An ideal voltmeter has infinite resistance.
$$S = \left(\frac{I_g}{I – I_g}\right) R_g \quad \text{(Ammeter Shunt)}$$
Compare & Contrast
✗ Coulomb’s Law (Electrostatics)
- Source is a scalar (charge $q$).
- Electric field produced is along the position vector connecting source to point.
- Independent of the angle.
- Acts on charges at rest or in motion.
✓ Biot-Savart Law (Magnetism)
- Source is a vector (current element $Id\vec{l}$).
- Magnetic field produced is perpendicular to both the current element and position vector.
- Highly dependent on angle ($\sin \theta$).
- Only acts on moving charges.
Remember
Both laws follow the inverse-square rule ($1/r^2$) and obey the principle of superposition, making them fundamental building blocks of electromagnetism.
Common Mistakes to Avoid
Mistake 1
Mixing up the Cross Product Order: The magnetic force is $\vec{F} = q(\vec{v} \times \vec{B})$. Many students accidentally write or calculate $q(\vec{B} \times \vec{v})$. Matrix cross multiplication is not commutative. Reversing them will give you the exact opposite direction for the force!
Mistake 2
Confusing $N$ and $n$ in Solenoids: In the formula $B = \mu_0 n I$, the $n$ represents the number of turns per unit length ($n = N/L$), NOT the total number of turns ($N$). Using the total number of turns instead of turn density is the most common mathematical error in solenoid problems.
Mistake 3
Misinterpreting $\alpha$ and $\theta$ in Torque: The formula $\tau = NIAB \sin \theta$ uses $\theta$, which is the angle between the Magnetic Field and the Area Vector (which is perpendicular to the coil’s plane). If a question gives you the angle $\alpha$ between the field and the plane of the coil itself, you must use $\tau = NIAB \cos \alpha$.
Exam Tips
Tip 1
To find the direction of the magnetic field at the center of a circular loop, use the Right-Hand Thumb Rule in reverse: Curl your fingers in the direction of the current flow around the loop, and your extended thumb points to the direction of the Magnetic Field!
Tip 2
When converting a galvanometer to an ammeter, the equivalent resistance of the new ammeter is practically equal to the shunt resistance $S$ (since $S \ll R_g$ and they are in parallel). This is a great shortcut to verify your calculations.
Did You Know
The Earth’s magnetic field protects us from deadly cosmic radiation (solar wind). As these charged particles hit the Earth’s magnetic field, the Lorentz force traps them into helical paths, funneling them towards the poles, which creates the beautiful Auroras (Northern Lights)!
Expected Exam Questions
SQ
Board Pattern Questions
Class 12 · Magnetism · CBSE Exam1
An electron and a proton enter a uniform magnetic field with the same kinetic energy, moving perpendicular to the field. Which one will have a larger radius of its circular path? [1 mark]
Answer
The Proton
📝
Explanation
The radius of the circular path is $r = \frac{mv}{qB}$. Since $v = \sqrt{2K/m}$ where $K$ is kinetic energy, substituting gives $r = \frac{\sqrt{2mK}}{qB}$. Since $K$, $q$, and $B$ are identical for both, $r \propto \sqrt{m}$. Because the mass of a proton is roughly 1836 times greater than an electron, the proton will trace a much larger radius.
2
A long straight wire carries a current of $35 \text{ A}$. What is the magnitude of the magnetic field at a distance of $20 \text{ cm}$ from the wire? [2 marks]
Answer
$3.5 \times 10^{-5} \text{ T}$
📝
Explanation
Given: $I = 35 \text{ A}$, $R = 20 \text{ cm} = 0.2 \text{ m}$, and we know $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$.
Using the formula for a long straight wire: $B = \frac{\mu_0 I}{2\pi R}$.
$B = \frac{4\pi \times 10^{-7} \times 35}{2\pi \times 0.2}$
$B = \frac{2 \times 10^{-7} \times 35}{0.2} = \frac{70 \times 10^{-7}}{0.2} = 350 \times 10^{-7} \text{ T} = 3.5 \times 10^{-5} \text{ T}$.
3
Increasing the current sensitivity of a galvanometer does not necessarily increase its voltage sensitivity. Explain why. [2 marks]
Answer
Because $V_s = I_s / R$. If $N$ doubles, $R$ also doubles, leaving $V_s$ unchanged.
📝
Explanation
Current sensitivity is $I_s = \frac{NBA}{k}$. Voltage sensitivity is $V_s = \frac{NBA}{kR}$, meaning $V_s = \frac{I_s}{R}$. The most common way to increase current sensitivity is to increase the number of turns ($N$). However, if you double $N$, the length of the wire doubles, which means the resistance ($R$) of the coil also doubles. Therefore, the ratio $I_s/R$ remains exactly the same, and voltage sensitivity does not increase.
Concept Map
Moving Charges & Magnetism connects to →
Electromagnetism
Magnetic Field Sources
Ampere’s Law
Lorentz Force
Cyclotron Mechanics
Magnetic Dipoles
Galvanometers
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