Introduction: Magnetic Field Lines & Magnetic Flux
A magnetic field is a region in space surrounding a magnet, current-carrying conductor, or moving charge within which magnetic forces can be detected. It is represented by the vector field quantity $\vec{B}$, known as magnetic flux density or magnetic induction. The SI unit of $\vec{B}$ is the Tesla (T) or $\text{Wb/m}^2$.
Magnetic Field Lines: Imaginary continuous visual lines drawn in a magnetic field such that the tangent at any point gives the direction of the magnetic field vector $\vec{B}$ at that point.
Fundamental Properties of Magnetic Field Lines:
- They form continuous, closed loops (unlike electrostatic field lines which begin on positive charges and end on negative charges).
- The tangent to a field line at any point gives the direction of the magnetic field $\vec{B}$.
- No two magnetic field lines can cross each other. If they did, two tangents could be drawn at the intersection point, implying two distinct directions for $\vec{B}$ at the same point, which is physically impossible.
- The density of field lines per unit cross-sectional area represents the strength of the magnetic field (denser lines represent a stronger field).
Magnetic Flux ($\Phi_B$)
Magnetic flux ($\Phi_B$) through a given surface area $A$ placed in a magnetic field is defined as the total number of magnetic field lines passing normally through that surface.
Mathematical Definition:
$$\Phi_B = \int \vec{B} \cdot d\vec{A} = \int B \cos\theta \, dA$$ For a uniform magnetic field $\vec{B}$ passing through a flat area $A$: $$\Phi_B = \vec{B} \cdot \vec{A} = B A \cos\theta$$ Where:- $\theta$ is the angle between the magnetic field $\vec{B}$ and the normal vector $\hat{n}$ to the surface.
- SI Unit: Weber ($\text{Wb}$), where $1 \text{ Wb} = 1 \text{ T} \cdot \text{m}^2 = 1 \text{ V} \cdot \text{s}$.
- Dimensional Formula: $[M^1 L^2 T^{-2} A^{-1}]$.
Oersted's Experiment, Outcome & Limitations
In 1820, Danish physicist Hans Christian Oersted discovered the fundamental connection between electricity and magnetism. Prior to this discovery, electricity and magnetism were treated as completely independent phenomena.
Experimental Setup & Procedure
A straight metallic conductor $AB$ is aligned parallel to the magnetic meridian (North-South direction) and placed directly over a pivoted magnetic needle. The wire is connected in series with a DC voltage source (battery), a rheostat, and a key $K$.
Key Observations:
- When no current flows ($I = 0$): The magnetic needle remains stationary along the Earth's North-South magnetic axis.
- When current flows from South to North ($S \to N$): The North pole ($N$-pole) of the magnetic needle immediately deflects towards the West. (Remembered via the SNOW Rule: Current from South to North Over needle deflects West).
- When current direction is reversed ($N \to S$): The $N$-pole of the needle deflects towards the East.
- When current magnitude $I$ is increased: The angle of deflection increases, demonstrating a stronger magnetic force.
Fundamental Outcome & Significance
Oersted's experiment established that moving electric charges (electric current) produce a magnetic field in surrounding space. This laid the foundation for electromagnetism.
Limitations of Oersted's Experiment
- It was purely qualitative in nature and did not provide a mathematical law or formula to calculate the exact magnitude of magnetic field $\vec{B}$.
- It applies only to steady continuous direct currents (DC).
- The observed deflection is influenced by the horizontal component of Earth's magnetic field ($B_H$), leading to systematic orientation errors.
Force on a Moving Charge in a Uniform Magnetic Field
When a point charge $q$ moves with velocity $\vec{v}$ through a region containing a uniform magnetic field $\vec{B}$, it experiences a side-wise deflecting force termed the Magnetic Lorentz Force ($\vec{F}_m$).
Lorentz Magnetic Force Vector Formula:
$$\vec{F}_m = q (\vec{v} \times \vec{B})$$ In scalar magnitude form: $$F_m = |q| v B \sin\theta$$ where $\theta$ is the angle between velocity vector $\vec{v}$ and magnetic field vector $\vec{B}$.Special Cases for Force Magnitude:
- Case 1 ($\theta = 0^\circ$ or $180^\circ$): If charge moves parallel or anti-parallel to field, $\sin\theta = 0 \implies F_m = 0$. Charge continues un-deflected.
- Case 2 ($\theta = 90^\circ$): If charge enters perpendicular to field, $\sin 90^\circ = 1 \implies F_{m,\text{max}} = q v B$. Force is maximum.
- Case 3 ($v = 0$): A stationary charge in a magnetic field experiences zero magnetic force ($F_m = 0$).
Direction of Force: Fleming's Left-Hand Rule
Stretch the thumb, forefinger, and middle finger of the left hand mutually perpendicular to each other:
- Forefinger: Points in the direction of Magnetic Field ($\vec{B}$).
- Middle Finger: Points in the direction of Motion of Positive Charge ($\vec{v}$).
- Thumb: Points in the direction of Magnetic Force ($\vec{F}_m$).
Motion of a Charged Particle in Uniform B-Field
Case A: Perpendicular Entry ($\theta = 90^\circ$) — Circular Motion
Since $\vec{F}_m \perp \vec{v}$ at all points, the magnetic force does zero work ($\vec{F}_m \cdot d\vec{r} = 0$) and changes only the direction of velocity without changing kinetic energy or speed. $\vec{F}_m$ acts as the necessary centripetal force:
$$\frac{m v^2}{r} = q v B \implies r = \frac{m v}{q B}$$Time period of revolution $T$ and Cyclotron Frequency $f$:
$$T = \frac{2\pi r}{v} = \frac{2\pi m}{q B}$$ $$f = \frac{1}{T} = \frac{q B}{2\pi m}$$Note: Time period and frequency are independent of particle velocity $v$ and radius $r$.
Case B: Oblique Entry ($\theta \neq 0^\circ, 90^\circ$) — Helical Motion
Resolve velocity into two components:
- $v_\parallel = v \cos\theta$ (parallel to $\vec{B}$, unchanged, causes linear displacement).
- $v_\perp = v \sin\theta$ (perpendicular to $\vec{B}$, causes circular trajectory of radius $r = \frac{m v \sin\theta}{q B}$).
The resulting trajectory is a Helix. The Pitch ($p$) is the linear distance advanced along the field per revolution:
$$p = v_\parallel \times T = (v \cos\theta) \cdot \left(\frac{2\pi m}{q B}\right)$$Force on a Current-Carrying Conductor in a Uniform B-Field
Since electric current in a conductor consists of moving free charge carriers, placing a current-carrying wire in an external magnetic field causes each drift charge to experience a Lorentz force. The microscopic sum of these forces results in a macroscopic force on the conductor.
Microscopic Derivation:
Consider a straight conductor segment of length $L$, cross-sectional area $A$, carrying steady current $I$.
- Let $n =$ number density of free electrons per unit volume.
- $v_d =$ drift velocity of electrons.
- $e =$ elementary electron charge ($1.6 \times 10^{-19} \text{ C}$).
Total number of free charge carriers in segment volume $V = A L$:
$$N = n \times (A L)$$Magnetic Lorentz force on a single free electron moving with velocity $\vec{v}_d$:
$$\vec{f}_e = -e (\vec{v}_d \times \vec{B})$$Total magnetic force $\vec{F}$ on the conductor segment:
$$\vec{F} = N \vec{f}_e = (n A L) [-e (\vec{v}_d \times \vec{B})] = - (n A e \vec{v}_d L) \times \vec{B}$$Recalling the microscopic current relation $I = n A e v_d$, and assigning vector length $\vec{L}$ in direction of conventional current:
$$\vec{F} = I (\vec{L} \times \vec{B})$$In scalar magnitude form:
$$F = I L B \sin\theta$$ where $\theta$ is the angle between wire length vector $\vec{L}$ and magnetic field $\vec{B}$.Force and Torque on a Rectangular Coil in Uniform B-Field
Consider a rectangular coil $PQRS$ of length $l$, width $b$, having $N$ turns carrying steady current $I$, placed in a uniform magnetic field $\vec{B}$. Let $\theta$ be the angle between the magnetic field vector $\vec{B}$ and the normal vector $\hat{n}$ to the plane of the coil.
Derivation of Torque ($\tau$):
- Forces on Horizontal Sides ($QR$ and $SP$): Equal in magnitude ($F = I b B \cos\theta$), opposite in direction, and act along the same line of action (axis of rotation). Thus, they cancel each other out with zero net force and zero net torque.
- Forces on Vertical Sides ($PQ$ and $RS$): Magnitude of force on each vertical side of length $l$: $$F_1 = F_2 = N I l B$$ These two forces are equal in magnitude, opposite in direction, but have different lines of action. Hence, they constitute a Deflecting Couple.
- Arm of Couple ($d$): Perpendicular distance between lines of action of $F_1$ and $F_2$: $$d = b \sin\theta$$
- Torque Magnitude ($\tau$): $$\tau = \text{Force} \times \text{Perpendicular distance} = (N I l B) \cdot (b \sin\theta)$$ Since surface area $A = l \times b$: $$\tau = N I A B \sin\theta$$
Vector Form of Torque:
$$\vec{\tau} = \vec{\mu} \times \vec{B}$$ where $\vec{\mu} = N I \vec{A}$ is the Magnetic Dipole Moment of the coil (SI Unit: $\text{A} \cdot \text{m}^2$).Work Done and Potential Energy
When the magnetic dipole rotates from initial angle $\theta_1$ to final angle $\theta_2$ in the B-field:
$$W = \int_{\theta_1}^{\theta_2} \tau \, d\theta = N I A B \int_{\theta_1}^{\theta_2} \sin\theta \, d\theta = - N I A B (\cos\theta_2 - \cos\theta_1)$$Magnetic Potential Energy ($U$):
$$U = -\vec{\mu} \cdot \vec{B} = -\mu B \cos\theta$$- Stable Equilibrium ($\theta = 0^\circ$): $U_{\text{min}} = -\mu B$ (Coil normal parallel to field).
- Unstable Equilibrium ($\theta = 180^\circ$): $U_{\text{max}} = +\mu B$ (Coil normal anti-parallel to field).
Moving Coil Galvanometer (MCG) & Applications
A Moving Coil Galvanometer is a sensitive electromagnetic instrument used to detect and measure extremely small electric currents (of the order of microamperes, $\mu\text{A}$).
Working Principle
When a current-carrying coil is suspended in a magnetic field, it experiences a torque ($\tau$). In a radial magnetic field, the plane of the coil is always parallel to the field lines ($\theta = 90^\circ$), resulting in a constant maximum deflecting torque.
Mathematical Theory & Equilibrium Equation:
1. Deflecting Torque ($\tau_d$):
$$\tau_d = N I A B \sin 90^\circ = N I A B$$2. Restoring Torque ($\tau_r$): Provided by the hairspring/phosphor-bronze strip twisted through angle $\theta$:
$$\tau_r = C \theta$$ where $C$ is the torsional restitution constant (restoring torque per unit twist).At equilibrium position ($\tau_d = \tau_r$):
$$N I A B = C \theta \implies I = \left(\frac{C}{N A B}\right) \theta$$ $$I = K \theta \implies I \propto \theta$$ where $K = \frac{C}{N A B}$ is the Galvanometer Constant. The linear relationship ($I \propto \theta$) enables a linear linear-spaced dial scale.Role of Concave Pole Pieces & Soft Iron Core
- Concave Cylindrical Pole Pieces: Produce a radial magnetic field so that magnetic field lines are always parallel to the coil plane ($\theta = 90^\circ$ always).
- Soft Iron Core: High relative permeability ($\mu_r \gg 1$) concentrates magnetic flux, dramatically increasing field strength $B$ and making the field strictly radial.
Sensitivities of MCG
1. Current Sensitivity ($S_I$)
Deflection produced per unit electric current:
$$S_I = \frac{\theta}{I} = \frac{N A B}{C} \quad [\text{rad/A or div/A}]$$2. Voltage Sensitivity ($S_V$)
Deflection produced per unit applied voltage:
$$S_V = \frac{\theta}{V} = \frac{\theta}{I R_g} = \frac{N A B}{C R_g} \quad [\text{div/V}]$$Applications: Conversion of MCG to Ammeter & Voltmeter
1. Conversion into Ammeter (To measure current in series)
A galvanometer is converted into an ammeter by connecting a low resistance (Shunt, $S$) in parallel with its coil.
Let $I_g =$ full-scale deflection current of MCG, $R_g =$ galvanometer coil resistance, $I =$ total current to be measured ($I > I_g$).
Since $R_g$ and $S$ are in parallel, potential difference across both branches is equal:
$$I_g R_g = (I - I_g) S \implies S = \frac{I_g R_g}{I - I_g}$$Effective Resistance of Ammeter ($R_A$):
$$R_A = \frac{R_g S}{R_g + S} \approx S \quad (\text{Very Low, ideal ammeter } R_A = 0)$$2. Conversion into Voltmeter (To measure voltage in parallel)
A galvanometer is converted into a voltmeter by connecting a high resistance ($R$) in series with its coil.
Let $V =$ total potential difference to be measured across range $0 - V$ volts.
Total resistance of voltmeter branch $= R_g + R$. Applying Ohm's Law:
$$V = I_g (R_g + R) \implies R = \frac{V}{I_g} - R_g$$Effective Resistance of Voltmeter ($R_V$):
$$R_V = R_g + R \quad (\text{Very High, ideal voltmeter } R_V = \infty)$$Hall Effect & Derivation of Hall Voltage ($V_H = \frac{BI}{ntq}$)
Hall Effect Definition: When a current-carrying conductor or semiconductor slab is placed in a transverse magnetic field, a potential difference is generated across its opposite faces perpendicular to both the direction of current and the magnetic field. This transverse voltage is known as the Hall Voltage ($V_H$).
Physical Mechanism & Derivation
Consider a rectangular semiconductor slab of length $l$, width $w$, thickness $t$, carrying steady current $I$ along the X-axis in a magnetic field $\vec{B}$ directed along the Z-axis.
Step-by-Step Derivation of Hall Voltage:
- Magnetic Lorentz Force ($F_M$): Charge carriers (charge $q$) drift with velocity $v_d$ along X-axis. They experience magnetic force along Y-axis: $$F_M = q v_d B$$
- Charge Accumulation & Transverse Electric Field ($E_H$): Accumulation of charges on the upper face creates an internal transverse electric field $E_H$. This exerts an opposing electrostatic force $F_E$: $$F_E = q E_H$$
- Dynamic Equilibrium Condition ($F_E = F_M$): $$q E_H = q v_d B \implies E_H = v_d B$$
- Relation between Hall Voltage $V_H$ and Electric Field $E_H$: Since transverse width is $w$: $$V_H = E_H \cdot w = (v_d B) w$$
- Substituting Microscopic Current Formula: Current $I = n A q v_d = n (w \cdot t) q v_d \implies v_d = \frac{I}{n w t q}$
- Final Hall Voltage Formula: $$V_H = \left(\frac{I}{n w t q}\right) B w \implies V_H = \frac{B I}{n t q}$$
Hall Coefficient ($R_H$):
$$R_H = \frac{1}{n q} \implies V_H = \frac{R_H B I}{t}$$Physical Significance of Hall Effect
- Determination of Charge Carrier Sign: Distinguishes whether conduction occurs via negative electrons ($n$-type) or positive holes ($p$-type).
- Measurement of Carrier Concentration ($n$): Directly calculates number density $n = \frac{I B}{V_H t q}$.
- Determination of Carrier Mobility ($\mu_e$): $\mu_e = \sigma |R_H|$, where $\sigma$ is electrical conductivity.
Hall Probe for Measuring Flux Density
A Hall Probe is a calibrated precision sensor based on the Hall effect, widely used in laboratory gaussmeters to measure unknown magnetic flux density $\vec{B}$.
Operational Procedure:
- A very thin semiconductor element (such as Indium Arsenide, $\text{InAs}$, or Gallium Arsenide, $\text{GaAs}$) with known thickness $t$ and carrier density $n$ is mounted at the tip of a non-magnetic probe rod.
- A calibrated constant control current $I$ is passed through the semiconductor sensor tip.
- The probe tip is oriented perpendicularly inside the unknown magnetic field $\vec{B}$ until maximum voltage reading is indicated on the digital display.
- The calibrated Hall voltage $V_H$ directly yields $B$: $$B = \frac{V_H \cdot n t q}{I} = \text{Constant} \times V_H$$
Biot-Savart Law & Applications
Biot-Savart Law is the fundamental law of magnetostatics that relates the magnetic field $d\vec{B}$ produced at a point in space to an infinitesimal current element $I d\vec{l}$.
Statement & Mathematical Formulation:
The magnitude of magnetic field $d B$ produced at point $P$ at distance $r$ from element $I d\vec{l}$ is:
- Directly proportional to current $I$.
- Directly proportional to element length $d l$.
- Directly proportional to $\sin\theta$ (where $\theta$ is angle between $d\vec{l}$ and position vector $\vec{r}$).
- Inversely proportional to the square of distance $r^2$.
Vector Form:
$$d\vec{B} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \hat{r})}{r^2} = \frac{\mu_0}{4\pi} \frac{I (d\vec{l} \times \vec{r})}{r^3}$$where $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A} = \text{Permeability of free space}$.
Applications of Biot-Savart Law
Application (i): Magnetic Field on the Axis of a Circular Current Coil
Consider a circular loop of radius $R$ carrying steady current $I$. We evaluate field $B$ at axial point $P$ at distance $x$ from coil center $O$.
Distance from current element $d l$ to axial point $P$: $r = \sqrt{R^2 + x^2}$.
Angle between element $d\vec{l}$ and position vector $\vec{r}$ is $\theta = 90^\circ$.
$$d B = \frac{\mu_0}{4\pi} \frac{I d l \sin 90^\circ}{r^2} = \frac{\mu_0}{4\pi} \frac{I d l}{R^2 + x^2}$$Resolving $d\vec{B}$ into axial component ($d B_x = d B \sin\phi$) and perpendicular component ($d B_\perp = d B \cos\phi$):
By symmetry, perpendicular components $\sum d B_\perp = 0$. Total field is sum of axial components:
$$B = \int d B \sin\phi = \int \left(\frac{\mu_0}{4\pi} \frac{I d l}{R^2 + x^2}\right) \left(\frac{R}{\sqrt{R^2 + x^2}}\right)$$ $$B = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} \int d l = \frac{\mu_0 I R}{4\pi (R^2 + x^2)^{3/2}} (2\pi R)$$For $N$ turns:
$$B_{\text{axis}} = \frac{\mu_0 N I R^2}{2 (R^2 + x^2)^{3/2}}$$Special Case: At center of coil ($x = 0$):
$$B_{\text{center}} = \frac{\mu_0 N I}{2 R}$$Application (ii): Field due to a Straight Current-Carrying Conductor
For a straight conductor at perpendicular distance $a$ with subtended angles $\phi_1$ and $\phi_2$ at the ends:
$$B = \frac{\mu_0 I}{4\pi a} (\sin\phi_1 + \sin\phi_2)$$For an infinitely long wire ($\phi_1 = \phi_2 = 90^\circ$):
$$B = \frac{\mu_0 I}{4\pi a} (\sin 90^\circ + \sin 90^\circ) = \frac{\mu_0 I}{2\pi a}$$Ampere's Circuital Law & Applications
Ampere's Circuital Law provides an elegant symmetry-based method to calculate magnetic fields, analogous to Gauss's Law in electrostatics.
Statement & Mathematical Integral:
The line integral of magnetic field vector $\vec{B}$ around any closed closed path (termed an Amperian Loop) in free space equals $\mu_0$ times the total net electric current $I_{\text{enc}}$ threading through the loop.
$$\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}$$Applications of Ampere's Law
1. Long Straight Current Wire
Choose a circular Amperian loop of radius $r$ centered on the wire. $\vec{B}$ is tangent to loop at all points ($\theta = 0^\circ$):
$$\oint \vec{B} \cdot d\vec{l} = B \oint d l = B (2\pi r) = \mu_0 I \implies B = \frac{\mu_0 I}{2\pi r}$$2. Straight Solenoid
Consider a rectangular Amperian loop $abcd$ of length $L$. Total turns enclosed $= n L$ (where $n = N/L$ is turns per unit length):
$$\oint_{abcd} \vec{B} \cdot d\vec{l} = \int_a^b \vec{B} \cdot d\vec{l} + \int_b^c \vec{B} \cdot d\vec{l} + \int_c^d \vec{B} \cdot d\vec{l} + \int_d^a \vec{B} \cdot d\vec{l}$$ $$\implies B L + 0 + 0 + 0 = \mu_0 (n L I) \implies B = \mu_0 n I$$3. Toroidal Solenoid (Toroid)
A toroid is an endless solenoid bent into a ring of mean radius $r$:
$$\oint \vec{B} \cdot d\vec{l} = B (2\pi r) = \mu_0 (N I) \implies B = \frac{\mu_0 N I}{2\pi r} = \mu_0 n I$$Field in interior empty space ($r < r_{\text{inner}}$) and exterior space ($r > r_{\text{outer}}$) is strictly zero ($B = 0$).
Force Between Two Parallel Conductors & Definition of Ampere
Consider two long, straight parallel conductors $C_1$ and $C_2$ placed in vacuum separated by distance $d$, carrying currents $I_1$ and $I_2$ respectively.
Step-by-Step Derivation of Force per Unit Length:
- Magnetic Field produced by wire 1 at wire 2 ($B_1$): $$B_1 = \frac{\mu_0 I_1}{2\pi d}$$ Directed perpendicularly into the plane of paper at wire 2 (by Right-Hand Thumb Rule).
- Magnetic Force on segment length $L$ of wire 2 ($F_{21}$): $$F_{21} = I_2 L B_1 \sin 90^\circ = I_2 L \left(\frac{\mu_0 I_1}{2\pi d}\right) = \frac{\mu_0 I_1 I_2 L}{2\pi d}$$
- Force per Unit Length ($f = F/L$): $$f = \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \quad [\text{N/m}]$$
Attraction vs Repulsion Rule
Parallel Currents (Same Direction)
Conductors ATTRACT each other.
Anti-Parallel Currents (Opposite Directions)
Conductors REPEL each other.
NEB Standard Board Exam Definition of 1 Ampere:
"One ampere is defined as that constant current which, if maintained in two straight parallel conductors of infinite length and negligible circular cross-section, placed 1 metre apart in vacuum, produces between them a force equal to $2 \times 10^{-7}$ newtons per metre of length."
Step-by-Step Solved Textbook Numericals
Example 1 (NEB 2078): Solenoid Field & Force
A solenoid $0.5 \text{ m}$ long has $500$ turns and carries a current of $2.5 \text{ A}$. Calculate the magnetic field strength at the center of the solenoid.
Given: $L = 0.5 \text{ m}$, $N = 500$, $I = 2.5 \text{ A}$, $\mu_0 = 4\pi \times 10^{-7} \text{ T}\cdot\text{m/A}$
Turns per unit length $n = \frac{N}{L} = \frac{500}{0.5} = 1000 \text{ turns/m}$
Formula: $B = \mu_0 n I = (4\pi \times 10^{-7}) \times 1000 \times 2.5$
Calculation: $B = 3.1416 \times 10^{-3} \text{ T} = 3.14 \text{ mT}$
Example 2 (NEB 2076): MCG Ammeter Conversion
A galvanometer has a coil resistance of $50 \, \Omega$ and shows full-scale deflection for a current of $2 \text{ mA}$. How can it be converted into an ammeter reading up to $5 \text{ A}$?
Given: $R_g = 50 \, \Omega$, $I_g = 2 \text{ mA} = 2 \times 10^{-3} \text{ A}$, $I = 5 \text{ A}$
Formula for Shunt Resistance: $S = \frac{I_g R_g}{I - I_g}$
$S = \frac{(2 \times 10^{-3}) \times 50}{5 - 0.002} = \frac{0.1}{4.998} \approx 0.020008 \, \Omega$
Result: Connect a shunt resistance of $0.020 \, \Omega$ in parallel with the galvanometer coil.
Example 3: Hall Effect Semiconductor Calculation
A ribbon of copper $1.0 \text{ mm}$ thick carries a current of $20 \text{ A}$ in a perpendicular magnetic field of $1.5 \text{ T}$. If carrier density $n = 8.49 \times 10^{28} \text{ m}^{-3}$, calculate the Hall voltage $V_H$.
Given: $t = 1.0 \text{ mm} = 10^{-3} \text{ m}$, $I = 20 \text{ A}$, $B = 1.5 \text{ T}$, $n = 8.49 \times 10^{28} \text{ m}^{-3}$, $q = 1.6 \times 10^{-19} \text{ C}$
Formula: $V_H = \frac{B I}{n t q}$
$V_H = \frac{1.5 \times 20}{(8.49 \times 10^{28}) \times 10^{-3} \times (1.6 \times 10^{-19})} = \frac{30}{1.3584 \times 10^7} \approx 2.21 \times 10^{-6} \text{ V} = 2.21 \, \mu\text{V}$
Conceptual Short-Answer Accordions
Q1: Why does a magnetic field do no work on a moving charged particle? ▼
The magnetic Lorentz force is given by $\vec{F} = q(\vec{v} \times \vec{B})$. By definition of vector cross product, $\vec{F}$ is always strictly perpendicular to the velocity vector $\vec{v}$ (and displacement $d\vec{r}$). Therefore, rate of work done $P = \vec{F} \cdot \vec{v} = 0$. Since work done is zero, the kinetic energy and speed of the particle remain constant.
Q2: Why is an ammeter always connected in series while a voltmeter is connected in parallel? ▼
An ammeter measures total current passing through a branch and has a very low resistance ($R_A \approx 0$). Connecting it in series ensures all current passes through it without significantly altering circuit resistance. Conversely, a voltmeter measures potential difference between two points and has very high resistance ($R_V \approx \infty$). Connecting it in parallel draws negligible current from the main circuit.
Q3: Why do two parallel wires carrying currents in opposite directions repel each other? ▼
The magnetic field $B_1$ produced by the first wire at the second wire interacts with the opposite current $I_2$. Applying Fleming's Left-Hand Rule to wire 2 shows that the resultant Lorentz force vector points away from wire 1, producing mutual repulsion.