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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 03 (Magnetic Effects of Current)

Chapter 04: Moving Charges and Magnetism

Master Revision Dossier • Biot-Savart, Ampere's Law, Helical Dynamics, Parallel Wires & Galvanometer

1. Definitive Formula & Magnetic Field Matrix

Concept Mathematical Formulation SI Unit / Dim Boundary Conditions & Notes
Biot-Savart Law $$d\vec{B} = \frac{\mu_0}{4\pi}\frac{I(d\vec{l}\times\hat{r})}{r^2} = \frac{\mu_0 I dl\sin\theta}{4\pi r^2}\hat{n}$$ $\text{Tesla (T)} = \text{Wb/m}^2$
$[\text{M}^1\text{T}^{-2}\text{A}^{-1}]$		 $\vec{B} \perp d\vec{l}$ and $\vec{B} \perp \vec{r}$. $\mu_0 = 4\pi \times 10^{-7}\text{ T}\cdot\text{m/A}$. Zero along element line ($\theta=0^\circ$).
Circular Loop Axis & Center $$B_{\text{axis}} = \frac{\mu_0 N I R^2}{2(R^2+x^2)^{3/2}}, \quad B_0 = \frac{\mu_0 N I}{2R}\text{}$$ $\text{Tesla (T)}$ Right-hand thumb rule gives axis sense. At far distance ($x \gg R$): $B \propto 1/x^3$ (magnetic dipole equivalent).
Ampere's Circuital Law $$\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enclosed}}\text{}$$ $\text{T}\cdot\text{m}$ Valid for any closed Amperian loop. Currents outside loop contribute to $\vec{B}$ but net integral cancels to zero.
Infinite Wire & Solenoid $$B_{\text{wire}} = \frac{\mu_0 I}{2\pi r}, \quad B_{\text{sol}} = \mu_0 n I\text{}$$ $\text{Tesla (T)}$ $n = N/L$ (turns per unit length). End of finite solenoid: $B_{\text{end}} = \frac{1}{2}\mu_0 n I$. Outside ideal solenoid $B \approx 0$.
Magnetic Lorentz Force $$\vec{F}_B = q(\vec{v}\times\vec{B}) \implies F_B = qvB\sin\theta\text{}$$ $\text{Newton (N)}$ $\vec{F}_B \perp \vec{v}$ always $\implies$ Work done $W = 0$, kinetic energy constant. Static charge feels $F_B = 0$.
Helical Motion Parameters $$r = \frac{mv\sin\theta}{qB}, \quad T = \frac{2\pi m}{qB}, \quad p = \frac{2\pi mv\cos\theta}{qB}$$ $r, p: \text{m}$
$T: \text{s}$		 Period $T$ and cyclotron frequency $f_c = \frac{qB}{2\pi m}$ are strictly independent of velocity $v$ and radius $r$.
Velocity Selector $$\vec{E} \perp \vec{B} \perp \vec{v} \implies v = \frac{E}{B}\text{}$$ $\text{m/s}$ $\vec{F}_E + \vec{F}_B = 0$. Undeflected trajectory condition is independent of particle mass $m$ and charge sign.
Force on Conductor $$\vec{F} = I(\vec{L}\times\vec{B}) = ILB\sin\theta\text{}$$ $\text{Newton (N)}$ Direction by Fleming's Left-Hand Rule. For any closed current loop in uniform field, $\oint d\vec{l} = 0 \implies \vec{F}_{\text{net}} = 0$.
Parallel Wires Force $$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}\text{}$$ $\text{N/m}$ Parallel currents attract; antiparallel repel. Defines SI Ampere: $2 \times 10^{-7}\text{ N/m}$ at $1\text{ m}$ separation in vacuum.
Torque on Loop & MCG $$\vec{\tau} = \vec{M}\times\vec{B}, \quad \tau = NIAB\sin\theta\text{}$$ $\text{N}\cdot\text{m}$ Radial field keeps $\theta = 90^\circ$ ($\sin 90^\circ = 1$). Restoring spring equilibrium: $NIAB = k\phi \implies \phi \propto I$.
Ammeter & Voltmeter $$S = \frac{I_g G}{I - I_g}, \quad R = \frac{V}{I_g} - G\text{}$$ $\text{Ohm (}\Omega\text{)}$ Shunt $S$ connected in parallel ($R_{\text{ammeter}} \to 0$); Multiplier $R$ connected in series ($R_{\text{voltmeter}} \to \infty$).

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 4.1: Magnetic Field on the Axis of a Circular Current Loop

1. Consider a circular loop of radius $R$ carrying steady current $I$. Point $P$ lies on its axis at distance $x$ from center $O$. Distance from any element $d\vec{l}$ to $P$ is $r = \sqrt{R^2 + x^2}$.

2. The element $d\vec{l}$ is tangential and perpendicular to position vector $\vec{r}$ ($\theta = 90^\circ$):

$$dB = \frac{\mu_0}{4\pi}\frac{I dl \sin 90^\circ}{r^2} = \frac{\mu_0}{4\pi}\frac{I dl}{R^2 + x^2}$$

3. Resolve $d\vec{B}$: Normal components $dB\cos\phi$ cancel pairwise ($\int dB\cos\phi = 0$). Axial components $dB\sin\phi$ add constructively:

$$B_{\text{net}} = \int dB\sin\phi = \int \left(\frac{\mu_0}{4\pi}\frac{I dl}{R^2 + x^2}\right) \left(\frac{R}{\sqrt{R^2 + x^2}}\right) = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \oint dl$$

4. Since $\oint dl = 2\pi R$, for a coil of $N$ tightly wound turns:

$$B_{\text{axis}} = \frac{\mu_0 N I R^2}{2(R^2 + x^2)^{3/2}} \quad \left(\text{At Center } x=0: B_0 = \frac{\mu_0 N I}{2R}\right)\text{}$$

Blueprint 4.2: Magnetic Field Inside an Ideal Long Straight Solenoid via ACL

1. Consider a long solenoid of turn density $n = N/L$ carrying current $I$. Construct a rectangular Amperian loop $abcd$ of length $L$ with $ab$ along the axis inside, $bc$ and $da$ perpendicular, and $cd$ outside.

2. Closed line integral evaluation:

$$\oint \vec{B}\cdot d\vec{l} = \int_a^b \vec{B}\cdot d\vec{l} + \int_b^c \vec{B}\cdot d\vec{l} + \int_c^d \vec{B}\cdot d\vec{l} + \int_d^a \vec{B}\cdot d\vec{l} = B L + 0 + 0 + 0 = B L$$

3. Total current enclosed by the loop is $I_{\text{enc}} = (n L) I$. By Ampere's Law $\oint \vec{B}\cdot d\vec{l} = \mu_0 I_{\text{enc}}$:

$$B L = \mu_0 (n L I) \implies B = \mu_0 n I\text{}$$

Blueprint 4.3: Force Between Parallel Conductors & Definition of 1 Ampere

1. Two long parallel conductors separated by distance $d$ carry currents $I_1$ and $I_2$. Wire 1 produces magnetic field at Wire 2: $B_1 = \frac{\mu_0 I_1}{2\pi d}$ (normal to Wire 2).

2. Magnetic force experienced by length $L$ of Wire 2: $F_{21} = I_2 L B_1 \sin 90^\circ = I_2 L \left(\frac{\mu_0 I_1}{2\pi d}\right)$.

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \quad (\vec{F}_{12} = -\vec{F}_{21} \text{ by Newton's 3rd Law})\text{}$$

3. Definition of 1 Ampere: If $I_1 = I_2 = 1\text{ A}$ and $d = 1\text{ m}$ in vacuum, force per unit length is strictly $\frac{F}{L} = \frac{(4\pi \times 10^{-7})(1)(1)}{2\pi(1)} = 2 \times 10^{-7}\text{ N/m}$.

Blueprint 4.4: Torque Balance and Conversions of Moving Coil Galvanometer

1. In a radial field produced by concave poles and soft iron core, $\theta = 90^\circ$ always. Deflecting torque: $\tau_{\text{def}} = N I A B$.

2. Restoring spring torque $\tau_{\text{rest}} = k \phi$. In equilibrium: $N I A B = k \phi \implies \phi = \left(\frac{N A B}{k}\right) I \implies \phi \propto I$.

3. Conversions: (i) Ammeter ($0$ to $I$): $S = \frac{I_g G}{I - I_g}$ in parallel. (ii) Voltmeter ($0$ to $V$): $R = \frac{V}{I_g} - G$ in series.

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Particle Orbit Radius Comparisons in Magnetic Field

Fixed Parameter Governing Radius Relation Proton ($p$) : Deuteron ($d$) : Alpha ($\alpha$) Physics Insight
Same Velocity ($v$) $r = \frac{mv}{qB} \implies r \propto \frac{m}{q}$ $1 : 2 : 2$ $m_\alpha = 4m_p, q_\alpha = 2q_p \implies m/q$ is equal for $d$ and $\alpha$.
Same Momentum ($p$) $r = \frac{p}{qB} \implies r \propto \frac{1}{q}$ $2 : 2 : 1 \quad (1 : 1 : 0.5)$ Radius is completely independent of particle mass $m$.
Same Kinetic Energy ($K$) $r = \frac{\sqrt{2mK}}{qB} \implies r \propto \frac{\sqrt{m}}{q}$ $1 : \sqrt{2} : 1$ Proton and alpha particle trace identical circular paths!
Same Accelerating Voltage ($V$) $r = \frac{1}{B}\sqrt{\frac{2mV}{q}} \implies r \propto \sqrt{\frac{m}{q}}$ $1 : \sqrt{2} : \sqrt{2}$ $K = qV$; substituted into the centripetal relation.

⚠️ Trap 2: Voltage vs Current Sensitivity Invariance

Increasing turns $N$ doubles current sensitivity ($I_s = \frac{NAB}{k} \propto N$). However, wire length doubles, so coil resistance also doubles ($R \propto N$). $$V_s = \frac{I_s}{R} = \frac{NAB}{k R} \implies \text{Constant!}$$ Increasing $I_s$ does NOT necessarily increase voltage sensitivity $V_s$.

⚠️ Trap 3: Work Done by Magnetic Force ($W_B = 0$)

Because $\vec{F}_B = q(\vec{v}\times\vec{B})$ is always strictly perpendicular to instantaneous displacement $d\vec{r}$: $$P = \vec{F}_B \cdot \vec{v} = 0 \implies W = \Delta K = 0$$ A static magnetic field can bend trajectories and alter momentum vectors, but can never alter particle kinetic energy or speed.

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Circular Coil Axial Field Drop Ratio Exemplar 01

At what distance on the axis of a circular current-carrying coil of radius $R$ does the magnetic field drop to $\frac{1}{8}\text{th}$ of its value at the center?

Solution: Field ratio on axis to center:

$$\frac{B_{\text{axis}}}{B_0} = \frac{R^3}{(R^2 + x^2)^{3/2}} = \left[\frac{R^2}{R^2 + x^2}\right]^{3/2} = \frac{1}{8} = \left(\frac{1}{2}\right)^3$$

Equating fractional powers:

$$\frac{R^2}{R^2 + x^2} = \frac{1}{4} \implies R^2 + x^2 = 4R^2 \implies x^2 = 3R^2 \implies x = \sqrt{3}R$$
Problem 2: Dual Galvanometer Conversion (Ammeter & Voltmeter) Exemplar 02

A galvanometer of coil resistance $G = 20\text{ }\Omega$ shows full-scale deflection for $I_g = 5\text{ mA}$. Calculate: (i) Shunt $S$ to convert it into an ammeter of range $0$ to $5\text{ A}$, (ii) Series resistance $R$ to convert it into a voltmeter of range $0$ to $20\text{ V}$.

Solution: Given $G = 20\text{ }\Omega$, $I_g = 5 \times 10^{-3}\text{ A}$.

1. Ammeter Shunt ($I = 5\text{ A}$):

$$S = \frac{I_g G}{I - I_g} = \frac{(5 \times 10^{-3})(20)}{5 - 0.005} = \frac{0.1}{4.995} \approx 0.02002\text{ }\Omega \approx 0.02\text{ }\Omega \quad (\text{in parallel})\text{}$$

2. Voltmeter Series Multiplier ($V = 20\text{ V}$):

$$R = \frac{V}{I_g} - G = \frac{20}{5 \times 10^{-3}} - 20 = 4000 - 20 = 3980\text{ }\Omega \quad (\text{in series})\text{}$$

5. Printable Pocket Flashcard Grid

Center Field of Coil

$B_0 = \frac{\mu_0 NI}{2R}$

Solenoid Core

$B = \mu_0 n I$

Parallel Wires

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

MCG Sensitivity

$I_s = \frac{NAB}{k}$