Physics: Class 12 (Book 1) — Chapter 4: Moving Charges and Magnetism

To represent three-dimensional vectors on a two-dimensional plane, the following standard conventions are adopted:

• Dot ($\odot$): Represents a current or a magnetic/electric field emerging perpendicularly out of the plane of the paper (towards the observer). This resembles the tip of an arrow pointed at you.
• Cross ($\otimes$): Represents a current or a magnetic/electric field going perpendicularly into the plane of the paper (away from the observer). This resembles the feathered tail of an arrow moving away from you.

Just as a static charge $Q$ acts as a source that produces an electric field $\vec{E}$, moving charges or currents produce a magnetic field $\vec{B}$. Both are vector fields and obey the principle of superposition , meaning the net field due to multiple sources is the vector sum of the individual fields:

$$\vec{B} = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 + \dots = \sum \vec{B}_i$$

When a point charge $q$ moves with velocity $\vec{v}$ in the presence of both an electric field $\vec{E}$ and a magnetic field $\vec{B}$, the total electromagnetic force experienced by the charge is called the Lorentz force :

$$\vec{F} = \vec{F}_{\text{electric}} + \vec{F}_{\text{magnetic}} = q \left[ \vec{E}(\vec{r}) + \vec{v} \times \vec{B}(\vec{r}) \right]$$

Key Features of the Magnetic Force ($\vec{F}_m = q [ \vec{v} \times \vec{B} ]$):

1. It depends directly on the charge $q$, the velocity vector $\vec{v}$, and the magnetic field vector $\vec{B}$. The force on a negative charge is opposite in direction to that on a positive charge.
2. The force is given by a vector cross product. Consequently, the magnetic force is always perpendicular to both the velocity $\vec{v}$ and the magnetic field $\vec{B}$. Its direction is determined by the right-hand screw rule.
3. The magnitude of the magnetic force is:$$F_m = q v B \sin\theta$$where $\theta$ is the angle between $\vec{v}$ and $\vec{B}$. The force vanishes (becomes zero) if the velocity is parallel ($\theta = 0^\circ$) or anti-parallel ($\theta = 180^\circ$) to the magnetic field.
4. The magnetic force is zero if the charge is stationary ($v = 0$). Only moving charges experience a magnetic force.

The SI unit of magnetic field $\vec{B}$ is the tesla ($T$), named after Nikola Tesla. Dimensionally, it is defined as:

$$[B] = \left[ \frac{F}{q v} \right] = \frac{\text{N}}{\text{C} \cdot \text{m/s}} = \text{N} \cdot \text{s} \cdot \text{C}^{-1} \cdot \text{m}^{-1} = \text{T}$$

A smaller, non-SI unit called the gauss ($G$) is also commonly used, where:

1 G = 10 ^-4 T

For reference, the Earth's magnetic field at the surface is approximately 3.6 &× 10 ^-5 T (or 0.36 G ).

We can extend the single-charge force analysis to a straight conducting rod of length $l$ and uniform cross-sectional area $A$ carrying a steady current $I$ in an external magnetic field $\vec{B}$:

1. Let $n$ be the number density of mobile charge carriers (electrons), each carrying charge $q$.
2. The total number of charge carriers in the rod is $n A l$.
3. For a steady current, each carrier moves with an average drift velocity $\vec{v}_d$.
4. The total force on these carriers is:$$\vec{F} = (n A l) q \vec{v}_d \times \vec{B}$$
5. Since the current density is $\vec{j} = n q \vec{v}_d$ and the total current is $I = |\vec{j}| A$, we can define a length vector $\vec{l}$ along the direction of the current:$$\vec{F} = I \vec{l} \times \vec{B}$$

For a wire of arbitrary shape, the total force is found by dividing the wire into infinitesimal linear elements $d\vec{l}$ and integrating:

$$\vec{F} = \int I d\vec{l} \times \vec{B}$$

When the velocity is perpendicular to the magnetic field, the magnetic force acts as a constant centripetal force , forcing the particle into a circular path of radius $r$:

$$\frac{m v^2}{r} = q v B \implies r = \frac{m v}{q B}$$

The angular frequency $\omega$ and frequency of rotation $\nu$ (often called the cyclotron frequency ) are given by:

$$\omega = \frac{v}{r} = \frac{q B}{m}$$

$$\nu = \frac{\omega}{2\pi} = \frac{q B}{2\pi m}$$

The time period for one complete revolution is:

$$T = \frac{1}{\nu} = \frac{2\pi m}{q B}$$

Crucial Observation: The frequency $\nu$ and time period $T$ are completely independent of the particle's speed $v$ and radius $r$. This independence is the fundamental operating principle of the cyclotron .

If the velocity $\vec{v}$ has a component parallel to the magnetic field ($\vec{v}_{\parallel} = v \cos\theta$) and a component perpendicular to the field ($\vec{v}_{\perp} = v \sin\theta$):

• The perpendicular component $\vec{v}_{\perp}$ produces circular motion in the plane perpendicular to $\vec{B}$, with radius:$$r = \frac{m v_{\perp}}{q B} = \frac{m v \sin\theta}{q B}$$
• The parallel component $\vec{v}_{\parallel}$ remains unaffected by the magnetic field and causes the particle to translate along the direction of the field.
• The combined motion is a helical path .

The linear distance traveled by the particle along the magnetic field direction during one period of revolution is called the pitch ($p$):

$$p = v_{\parallel} T = v \cos\theta \left( \frac{2\pi m}{q B} \right) = \frac{2\pi m v_{\parallel}}{q B}$$

An elegant relation exists between the permittivity of free space $\epsilon_0$, the permeability of free space $\mu_0$, and the speed of light in vacuum $c$:

$$\epsilon_0 \mu_0 = (4\pi \epsilon_0) \left( \frac{\mu_0}{4\pi} \right) = \left( \frac{1}{9 \times 10^9} \right) \times (10^{-7}) = \frac{1}{9 \times 10^{16}} = \frac{1}{c^2}$$

$$\epsilon_0 \mu_0 = \frac{1}{c^2}$$

At the center of the circular loop, the axial distance $x = 0$. Substituting this into the axial field equation yields:

$$\vec{B}_0 = \frac{\mu_0 I}{2R} \hat{i}$$

For a coil consisting of $N$ closely wound turns, the magnetic field is multiplied by $N$:

$$\vec{B}_0 = \frac{\mu_0 N I}{2R} \hat{i}$$

Consider an infinitely long, straight wire carrying a steady current $I$. We choose a circular Amperian loop of radius $r$ centered on the wire and perpendicular to it:

1. By symmetry, the magnetic field $\vec{B}$ is constant in magnitude at all points on the circle and is directed tangentially to the circle at every point.
2. Therefore, the angle between $\vec{B}$ and $d\vec{l}$ is $0^\circ$:$$\oint \vec{B} \cdot d\vec{l} = \oint B dl \cos(0^\circ) = B \oint dl = B (2\pi r)$$
3. Apply Ampere's law:$$B (2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$$

This result shows that the magnetic field possesses cylindrical symmetry and decreases inversely with the distance $r$ from the wire.

For a long straight wire of circular cross-section of radius $a$ carrying a uniform current $I$:

• Region 1: Outside the wire ($r > a$): The enclosed current is the total current $I$.$$B = \frac{\mu_0 I}{2\pi r} \implies B \propto \frac{1}{r}$$
• Region 2: Inside the wire ($r < a$): Assuming uniform current density, the current enclosed by an Amperian loop of radius $r$ is:$$I_e = I \left( \frac{\pi r^2}{\pi a^2} \right) = I \frac{r^2}{a^2}$$Applying Ampere's law:$$B (2\pi r) = \mu_0 I_e = \mu_0 I \frac{r^2}{a^2} \implies B = \left( \frac{\mu_0 I}{2\pi a^2} \right) r \implies B \propto r$$

Consider a rectangular Amperian loop $abcd$ of length $h$ where side $ab$ lies inside the solenoid parallel to its axis, and side $cd$ lies outside:

1. Evaluate the line integral $\oint \vec{B} \cdot d\vec{l}$ along the four segments:$$\oint \vec{B} \cdot d\vec{l} = \int_a^b \vec{B} \cdot d\vec{l} + \int_b^c \vec{B} \cdot d\vec{l} + \int_c^d \vec{B} \cdot d\vec{l} + \int_d^a \vec{B} \cdot d\vec{l}$$
2. Outside the solenoid, the field is zero, so $\int_c^d \vec{B} \cdot d\vec{l} = 0$.
3. Along the transverse segments $bc$ and $da$, the magnetic field is perpendicular to the path ($\vec{B} \perp d\vec{l}$), so $\vec{B} \cdot d\vec{l} = 0$.
4. Inside the solenoid, the field is parallel to the axis, so:$$\int_a^b \vec{B} \cdot d\vec{l} = B h$$
5. Let $n$ be the number of turns per unit length. The total number of turns enclosed by the loop of length $h$ is $n h$. The total enclosed current is:$$I_e = n h I$$
6. Applying Ampere's law:$$B h = \mu_0 (n h I) \implies B = \mu_0 n I$$

• Parallel currents (flowing in the same direction): Attract each other.
• Antiparallel currents (flowing in opposite directions): Repel each other.

Note: This is opposite to electrostatics, where like charges repel and unlike charges attract.

The force between parallel currents is used to define the base SI unit of current, the ampere ($A$):

"The ampere is that steady current which, if maintained in two extremely long, straight, parallel conductors of negligible circular cross-section, and placed one metre apart in vacuum, would produce on each of these conductors a force equal to 2 &× 10 ^-7 newtons per metre of length."

Consider a rectangular loop carrying a steady current $I$ placed in a uniform magnetic field $\vec{B}$:

1. The net force on the loop in a uniform magnetic field is always zero ($\vec{F}_{\text{net}} = 0$).
2. However, the forces on opposite arms of the loop can create a couple, producing a net torque ($\vec{\tau}$).
3. If the normal to the plane of the loop makes an angle $\theta$ with the magnetic field $\vec{B}$, the magnitude of the torque is:$$\tau = I A B \sin\theta$$where $A$ is the area of the loop.

We define the magnetic dipole moment ($\vec{m}$) of a current loop as:

$$\vec{m} = I \vec{A}$$

For a coil with $N$ turns:

$$\vec{m} = N I \vec{A}$$

The direction of the area vector $\vec{A}$ (and thus $\vec{m}$) is given by the right-hand thumb rule: curl the fingers of your right hand in the direction of the current, and your thumb points in the direction of $\vec{m}$.

Using this definition, the torque can be written compactly as a vector cross product:

$$\vec{\tau} = \vec{m} \times \vec{B}$$

This is directly analogous to the electrostatic torque on an electric dipole ($\vec{\tau} = \vec{p} \times \vec{E}$).

The MCG consists of a coil with $N$ turns suspended in a uniform radial magnetic field produced by cylindrical pole pieces and a soft iron core. The radial field ensures that the plane of the coil is always parallel to the magnetic field lines ($\theta = 90^\circ$), making the deflecting torque maximum and constant:

$$\tau_{\text{deflecting}} = N I A B$$

This deflecting torque is balanced by the restoring torque of a suspension spring with torsional constant $k$:

$$\tau_{\text{restoring}} = k \phi$$

In equilibrium:

$$k \phi = N I A B \implies \phi = \left( \frac{N A B}{k} \right) I$$

where $\phi$ is the steady angular deflection of the pointer.

• Current Sensitivity ($I_s$): Deflection per unit current.$$$I_s = \frac{\phi}{I} = \frac{N A B}{k}$$
• Voltage Sensitivity ($V_s$): Deflection per unit voltage.$$$V_s = \frac{\phi}{V} = \frac{\phi}{I R} = \frac{N A B}{k R}$$where $R$ is the resistance of the galvanometer coil.

Note: Doubling the number of turns ($N \to 2N$) doubles the current sensitivity, but since the resistance also doubles ($R \to 2R$), the voltage sensitivity remains unchanged.