Chapter 19: Moving Charges and Magnetism

Introduction

Key Points

• In 1820, Hans Christian Oersted discovered that electric current in a wire produces a magnetic field affecting a compass needle.
• The magnetic field lines form concentric circles around the wire; direction depends on current direction.
• A dot (⊙) indicates current/magnetic field coming out of the paper; a cross (⊗) indicates going into the paper.
• James Maxwell unified electricity and magnetism, showing light as an electromagnetic wave.
• Radio waves were discovered by Hertz and produced by J.C. Bose and G. Marconi.
• This chapter covers magnetic forces on moving charges and current-carrying wires, magnetic fields from currents, and galvanometer operation.

Solved Examples

• Ex 1: Determine the direction of magnetic field around a wire carrying current out of the paper.
  • Use right-hand thumb rule: thumb points out of paper, fingers curl in direction of magnetic field.

Practice Set

• Q1: What does a dot (⊙) symbol represent in magnetic field diagrams?
  • A) Field into paper
  • B) Field out of paper
  • C) No field
  • D) Field parallel to paper
• Q2: Who unified electricity and magnetism?
  • A) Oersted
  • B) Maxwell
  • C) Hertz
  • D) Marconi
• Q3: The magnetic field lines around a current-carrying wire are:
  • A) Straight lines
  • B) Radial lines
  • C) Concentric circles
  • D) Random

Answer Key

• Q1: B - Dot (⊙) indicates magnetic field or current coming out of the paper.
• Q2: B - James Maxwell unified electricity and magnetism.
• Q3: C - Magnetic field lines form concentric circles around the wire.

Memory Aids

• Right-hand thumb rule: Thumb = current direction, fingers = magnetic field direction.
• Dot (⊙) = out of paper; Cross (⊗) = into paper.

Magnetic Force

Sources and Fields

• Moving charges and electric currents produce magnetic field \( \vec{B} \), a vector quantity defined at every point in space.
• Principle of Superposition: Total magnetic field is vector sum of fields from individual sources.
• Lorentz force on a charge \( q \) moving with velocity \( \vec{v} \) in electric \( \vec{E} \) and magnetic \( \vec{B} \) fields: \[ \vec{F} = q(\vec{E} + \vec{v} \times \vec{B}) = \vec{F}_e + \vec{F}_m \]
• Magnetic force magnitude: \[ |\vec{F}_m| = q v B \sin \theta \] where \( \theta \) is angle between \( \vec{v} \) and \( \vec{B} \).
• Magnetic force depends on charge \( q \), velocity \( \vec{v} \), and magnetic field \( \vec{B} \).
• Force is maximum when \( \vec{v} \perp \vec{B} \), zero when parallel or antiparallel.
• Only moving charges experience magnetic force.
• Direction of magnetic force on positive charge given by right-hand rule: point fingers along \( \vec{v} \), curl toward \( \vec{B} \), thumb points to \( \vec{F}_m \).
• Negative charges experience force in opposite direction.
• Unit of magnetic field (Tesla, T): \[ [B] = \frac{N}{C \cdot m/s} = \text{Tesla (T)} \] 1 gauss = \( 10^{-4} \) Tesla.

Magnetic Force on Current-Carrying Conductor

• Charge in conductor: \( q = n A l e \), where \( n \) = charge carrier density, \( A \) = cross-sectional area, \( l \) = length, \( e \) = charge.
• Drift velocity: \( \vec{v}_d \).
• Magnetic force on charges: \[ \vec{F}_m = (n A l) e (\vec{v}_d \times \vec{B}) = (\vec{J} A l) \times \vec{B} = I (\vec{l} \times \vec{B}) \] where \( \vec{J} = n e \vec{v}_d \) is current density, \( I \) is current, \( \vec{l} \) is length vector.
• For arbitrary wire shape: \[ \vec{F} = I \sum (d\vec{l}_k \times \vec{B}) \]

Solved Examples

• Ex 1: Calculate magnetic force on a wire of length 0.5 m carrying 3 A current perpendicular to a 0.2 T magnetic field.
  • Use formula: \( F = I l B \sin 90^\circ = 3 \times 0.5 \times 0.2 = 0.3 \text{ N} \)
• Ex 2: Find direction of force on a positive charge moving right in a magnetic field into the page.
  • Use right-hand rule: velocity right, magnetic field into page, force is upward.

Practice Set

• Q1: Magnetic force on a charge is zero when:
  • A) \( \vec{v} \perp \vec{B} \)
  • B) \( \vec{v} \parallel \vec{B} \)
  • C) Charge is moving
  • D) Magnetic field is zero
• Q2: Unit of magnetic field Tesla (T) is equivalent to:
  • A) N/C
  • B) N s C^-1 m^-1
  • C) V/m
  • D) Wb/m²
• Q3: A wire of length 0.4 m carries 5 A current in a magnetic field of 0.3 T at 60° angle. Find magnetic force magnitude.
  • A) 0.6 N
  • B) 0.52 N
  • C) 1.0 N
  • D) 0.3 N

Answer Key

• Q1: B - Force is zero when velocity is parallel or antiparallel to magnetic field.
• Q2: B - Tesla unit is N s C^-1 m^-1.
• Q3: A - \( F = I l B \sin \theta = 5 \times 0.4 \times 0.3 \times \sin 60^\circ = 0.6 \text{ N} \).

Memory Aids

• Magnetic force formula: \( F = q v B \sin \theta \).
• Force on wire: \( \vec{F} = I (\vec{l} \times \vec{B}) \).
• Right-hand rule for force direction on positive charges and current.

Motion in a Magnetic Field

Velocity Parallel or Anti-Parallel to Magnetic Field (\(\theta=0\) or \(\pi\))

• Magnetic force \(\vec{F} = q\vec{v} \times \vec{B} = 0\) since velocity and magnetic field are parallel.
• Particle moves in a straight line with constant speed.

Velocity Perpendicular to Magnetic Field (\(\theta=\pi/2\))

• Magnetic force \(F_B = qvB\) acts perpendicular to velocity and magnetic field.
• Force acts as centripetal force: \(qvB = \frac{mv^2}{r}\).
• Radius of circular path: \(r = \frac{mv}{qB} = \frac{p}{qB} = \frac{\sqrt{2mK}}{qB}\), where \(p\) is momentum and \(K\) is kinetic energy.
• Time period of revolution: \(T = \frac{2\pi m}{qB}\).
• Frequency: \(f = \frac{qB}{2\pi m}\).
• Angular speed: \(\omega = \frac{qB}{m}\).
• Time period and frequency depend only on magnetic field \(B\) and charge-to-mass ratio \(\frac{q}{m}\), independent of speed and radius.

Velocity at an Angle to Magnetic Field (\(0 < \theta < \pi/2\))

• Velocity components: parallel \(v_{\parallel} = v \cos \theta\), perpendicular \(v_{\perp} = v \sin \theta\).
• Parallel component remains constant; perpendicular component causes circular motion.
• Particle moves in a helical path around magnetic field lines.
• Radius of circular motion: \(r = \frac{mv_{\perp}}{qB} = \frac{mv \sin \theta}{qB}\).
• Time period of revolution: \(T = \frac{2\pi m}{qB}\) (same as perpendicular case).
• Pitch of helix (distance moved along field in one revolution): \(P = v_{\parallel} T = \frac{2\pi m v \cos \theta}{qB}\).

Deviation of Charged Particle in Magnetic Field

• Particle of charge \(q\) and mass \(m\) enters uniform magnetic field \(B\) of width \(d\).
• Moves in circular path of radius \(R = \frac{mv}{qB}\) inside field.
• Enters at point P, exits at Q, then moves straight.
• Angle of deviation \(\theta\) given by \(\sin \theta = \frac{d}{R}\) for \(d \leq R\).
• If \(d > R\), deviation angle \(\theta = \pi\) (particle reverses direction).

Solved Examples

Ex 1

• Calculate radius of circular path for electron (mass \(9.11 \times 10^{-31}\) kg, charge \(1.6 \times 10^{-19}\) C) moving at \(2 \times 10^6\) m/s in magnetic field \(0.01\) T.
• Use \(r = \frac{mv}{qB} = \frac{9.11 \times 10^{-31} \times 2 \times 10^6}{1.6 \times 10^{-19} \times 0.01} = 1.14 \times 10^{-3}\) m.

Ex 2

• Find time period of revolution for proton (mass \(1.67 \times 10^{-27}\) kg, charge \(1.6 \times 10^{-19}\) C) in magnetic field \(0.5\) T.
• Use \(T = \frac{2\pi m}{qB} = \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.5} = 1.31 \times 10^{-7}\) s.

Ex 3

• A particle moves with velocity at \(30^\circ\) to magnetic field \(0.2\) T. Mass \(2 \times 10^{-27}\) kg, charge \(1.6 \times 10^{-19}\) C, speed \(3 \times 10^6\) m/s. Find radius of helical path.
• Calculate \(v_{\perp} = v \sin 30^\circ = 1.5 \times 10^6\) m/s.
• Radius \(r = \frac{mv_{\perp}}{qB} = \frac{2 \times 10^{-27} \times 1.5 \times 10^6}{1.6 \times 10^{-19} \times 0.2} = 9.38 \times 10^{-3}\) m.

Practice Set

Q1

• A charged particle moves parallel to a magnetic field. What is the path of the particle?
• A. Circular
• B. Helical
• C. Straight line
• D. Elliptical

Q2

• Radius of circular path of a charged particle in magnetic field depends on:
• A. Speed and magnetic field only
• B. Mass, speed, charge, and magnetic field
• C. Charge and magnetic field only
• D. Mass and charge only

Q3

• A proton moves with velocity \(4 \times 10^6\) m/s at \(60^\circ\) to a magnetic field of \(0.3\) T. Find the pitch of the helical path. (Mass of proton = \(1.67 \times 10^{-27}\) kg, charge = \(1.6 \times 10^{-19}\) C)
• A. \(2.2 \times 10^{-2}\) m
• B. \(4.4 \times 10^{-2}\) m
• C. \(1.1 \times 10^{-2}\) m
• D. \(3.3 \times 10^{-2}\) m

Answer Key

• Q1: C. No magnetic force acts when velocity is parallel; path is straight.
• Q2: B. Radius \(r = \frac{mv}{qB}\) depends on mass, speed, charge, and magnetic field.
• Q3: D. Pitch \(P = v_{\parallel} T = v \cos \theta \times \frac{2\pi m}{qB} = 4 \times 10^6 \times 0.5 \times \frac{2\pi \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.3} = 3.3 \times 10^{-2}\) m.

Memory Aids

• Force zero if velocity \(\parallel\) magnetic field.
• Radius \(r = \frac{mv}{qB}\) for circular motion.
• Time period \(T = \frac{2\pi m}{qB}\) independent of speed.
• Helical path if velocity at angle; pitch \(P = v \cos \theta \times T\).
• Deviation angle \(\sin \theta = \frac{d}{R}\) for particle in magnetic field region.

Magnetic Field due to a Current Element, Biot-Savart Law

Explanation

• A current element \( I d\vec{l} \) produces a magnetic field \( d\vec{B} \) at point P located at distance \( r \).
• Magnitude: \( |d\vec{B}| = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2} \), where \( \theta \) is angle between \( d\vec{l} \) and \( \vec{r} \).
• Vector form: \( d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \vec{r}}{r^3} \).
• Direction of \( d\vec{B} \) is perpendicular to the plane containing \( d\vec{l} \) and \( \vec{r} \) (right-hand rule).
• Permeability of free space: \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T \cdot m/A} \).
• Magnetic field is a vector field produced by vector source \( I d\vec{l} \), unlike electric field from scalar charge.
• Magnetic field depends on angle \( \theta \); zero along direction of \( d\vec{l} \).
• Relation: speed of light \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} = 3 \times 10^8 \ \mathrm{m/s} \).

Solved Examples

• Ex 1: Calculate \( dB \) due to a current element \( I=2A \), length \( dl=0.01m \), at distance \( r=0.1m \), angle \( \theta=90^\circ \).
  • Use formula: \( dB = \frac{\mu_0}{4\pi} \frac{I dl \sin \theta}{r^2} \)
  • Substitute: \( dB = 10^{-7} \times \frac{2 \times 0.01 \times 1}{0.01} = 2 \times 10^{-7} \ \mathrm{T} \)

Practice Set

• Q1: The magnetic field \( dB \) due to a current element is zero when angle between \( d\vec{l} \) and \( \vec{r} \) is:
  • A) 0°
  • B) 45°
  • C) 90°
  • D) 180°
• Q2: The permeability of free space \( \mu_0 \) is:
  • A) \( 4\pi \times 10^{-7} \ \mathrm{T \cdot m/A} \)
  • B) \( 8.85 \times 10^{-12} \ \mathrm{F/m} \)
  • C) \( 3 \times 10^8 \ \mathrm{m/s} \)
  • D) \( 1.6 \times 10^{-19} \ \mathrm{C} \)
• Q3: The vector form of Biot-Savart law is:
  • A) \( d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \cdot \vec{r}}{r^3} \)
  • B) \( d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \vec{r}}{r^3} \)
  • C) \( d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l}}{r^2} \)
  • D) \( d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{\vec{r}}{r^2} \)

Answer Key

• Q1: A) 0° - Magnetic field is zero when \( \sin \theta = 0 \), i.e., \( \theta = 0^\circ \) or \( 180^\circ \).
• Q2: A) \( 4\pi \times 10^{-7} \ \mathrm{T \cdot m/A} \) - This is the permeability of free space.
• Q3: B) \( d\vec{B} = \frac{\mu_0 I}{4\pi} \frac{d\vec{l} \times \vec{r}}{r^3} \) - Vector form involves cross product.

Memory Aids

• Biot-Savart law formula: \( dB \propto I dl \sin \theta / r^2 \).
• Direction: Use right-hand thumb rule for \( d\vec{l} \times \vec{r} \).
• \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T \cdot m/A} \).
• Magnetic field zero along current element direction.

Magnetic Field due to Current Carrying Straight Conductor

Explanation

• Consider a straight wire AB carrying current \( I \), point P at perpendicular distance \( d \).
• Magnetic field contribution from element \( dl \) making angle \( 90^\circ + \theta \) with \( \vec{r} \): \( dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(90^\circ + \theta)}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl \cos \theta}{r^2} \).
• Using geometry: \( dl = d \sec^2 \theta d\theta \), \( r = d \sec \theta \).
• Total magnetic field at P: \[ B = \int dB = \frac{\mu_0 I}{4\pi d} \int_{\theta_1}^{\theta_2} \cos \theta d\theta = \frac{\mu_0 I}{4\pi d} (\sin \theta_2 - \sin \theta_1) \]
• For semi-infinite wire: \( \theta_1 = 0, \theta_2 = \frac{\pi}{2} \), so \( B = \frac{\mu_0 I}{4\pi d} \times 1 = \frac{\mu_0 I}{4\pi d} \).
• For infinite wire: \( \theta_1 = -\frac{\pi}{2}, \theta_2 = \frac{\pi}{2} \), so \( B = \frac{\mu_0 I}{2\pi d} \).
• Magnetic field decreases inversely with distance \( d \): \( B \propto \frac{1}{d} \).

Solved Examples

• Ex 1: Calculate magnetic field at point P due to a straight wire of length \( l=0.5m \), current \( I=3A \), distance \( d=0.1m \), angles \( \theta_1=30^\circ, \theta_2=90^\circ \).
  • Use formula: \( B = \frac{\mu_0 I}{4\pi d} (\sin \theta_2 - \sin \theta_1) \)
  • Calculate: \( B = 10^{-7} \times \frac{3}{0.1} (1 - 0.5) = 1.5 \times 10^{-6} \ \mathrm{T} \)
• Ex 2: Magnetic field due to infinite wire at distance \( d=0.2m \), current \( I=5A \).
  • Use formula: \( B = \frac{\mu_0 I}{2\pi d} \)
  • Calculate: \( B = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.2} = 5 \times 10^{-6} \ \mathrm{T} \)

Practice Set

• Q1: Magnetic field at point P due to semi-infinite wire carrying current \( I \) at distance \( d \) is:
  • A) \( \frac{\mu_0 I}{2\pi d} \)
  • B) \( \frac{\mu_0 I}{4\pi d} \)
  • C) \( \frac{\mu_0 I}{\pi d} \)
  • D) \( \frac{\mu_0 I}{8\pi d} \)
• Q2: For a finite wire, magnetic field depends on angles \( \theta_1 \) and \( \theta_2 \) as:
  • A) \( \sin \theta_2 - \sin \theta_1 \)
  • B) \( \cos \theta_2 - \cos \theta_1 \)
  • C) \( \tan \theta_2 - \tan \theta_1 \)
  • D) \( \sec \theta_2 - \sec \theta_1 \)
• Q3: Magnetic field at 0.1 m from infinite wire carrying 4 A current is:
  • A) \( 8 \times 10^{-6} \ \mathrm{T} \)
  • B) \( 2 \times 10^{-6} \ \mathrm{T} \)
  • C) \( 4 \times 10^{-6} \ \mathrm{T} \)
  • D) \( 1 \times 10^{-6} \ \mathrm{T} \)

Answer Key

• Q1: B) \( \frac{\mu_0 I}{4\pi d} \) - Semi-infinite wire formula.
• Q2: A) \( \sin \theta_2 - \sin \theta_1 \) - Magnetic field depends on sine difference of angles.
• Q3: C) \( 4 \times 10^{-6} \ \mathrm{T} \) - Using \( B = \frac{\mu_0 I}{2\pi d} \).

Memory Aids

• Magnetic field \( B \propto \frac{I}{d} \) for straight wire.
• Use \( B = \frac{\mu_0 I}{4\pi d} (\sin \theta_2 - \sin \theta_1) \) for finite wire.
• Infinite wire: \( B = \frac{\mu_0 I}{2\pi d} \).
• Angles \( \theta_1, \theta_2 \) measured from wire ends to point P.

Magnetic Field on the Axis of a Circular Current Loop

Concepts and Formulas

• Magnetic field due to small element \( d\vec{l} \) carrying current \( I \):
  \( |d\vec{B}| = \frac{\mu_0}{4\pi} \frac{I |d\vec{l} \times \vec{r}|}{r^3} = \frac{\mu_0}{4\pi} \frac{I dl}{r^2} = \frac{\mu_0}{4\pi} \frac{I dl}{x^2 + R^2} \)
• Due to symmetry, only the x-component \( dB_x = dB \cos \theta = \frac{\mu_0 I dl}{4\pi} \frac{R}{(x^2 + R^2)^{3/2}} \) contributes to net field.
• Net magnetic field at point \( P \) on axis:
  \( \vec{B} = B_x \hat{i} = \frac{\mu_0 I R}{4\pi (x^2 + R^2)^{3/2}} \int dl \hat{i} \)
• Since \( \int dl = 2\pi R \), total field:
  \( \vec{B} = \frac{\mu_0 I R^2}{2 (x^2 + R^2)^{3/2}} \hat{i} \)
• Direction by right-hand thumb rule: Curl fingers in current direction, thumb points magnetic field direction.
• At center of loop (\( x=0 \)):
  \( B = \frac{\mu_0 I}{2 R} \)
• For current carrying arc subtending angle \( \theta \):
  \( B = \frac{\mu_0 I \theta}{4 \pi R} \)
• For \( x \gg R \), magnetic field:
  \( B = \frac{\mu_0 I R^2}{2 x^3} \) (dipole field, \( B \propto \frac{1}{x^3} \))

Solved Examples

• Ex 1: Find magnetic field at center of loop with radius 0.1 m carrying 5 A current.
  Solution:
  \( B = \frac{\mu_0 I}{2 R} = \frac{4\pi \times 10^{-7} \times 5}{2 \times 0.1} = 3.14 \times 10^{-5} \text{ T} \)
• Ex 2: Magnetic field at point 0.2 m from center on axis of loop with radius 0.1 m and current 3 A.
  Solution:
  \( B = \frac{\mu_0 I R^2}{2 (x^2 + R^2)^{3/2}} = \frac{4\pi \times 10^{-7} \times 3 \times 0.01}{2 (0.04 + 0.01)^{3/2}} = 1.7 \times 10^{-6} \text{ T} \)

Practice Set

• Q1: Magnetic field at center of circular loop of radius 0.05 m carrying 2 A current is:
  A) \( 8 \times 10^{-6} \text{ T} \)
  B) \( 2.5 \times 10^{-5} \text{ T} \)
  C) \( 1.26 \times 10^{-5} \text{ T} \)
  D) \( 5 \times 10^{-6} \text{ T} \)
• Q2: Magnetic field at point on axis 0.1 m from center of loop radius 0.05 m, current 4 A:
  A) \( 3.2 \times 10^{-6} \text{ T} \)
  B) \( 1.6 \times 10^{-6} \text{ T} \)
  C) \( 6.4 \times 10^{-6} \text{ T} \)
  D) \( 8 \times 10^{-6} \text{ T} \)
• Q3: Magnetic field at center due to arc of radius 0.1 m, current 3 A, subtending \( 90^\circ \):
  A) \( 4.77 \times 10^{-6} \text{ T} \)
  B) \( 9.55 \times 10^{-6} \text{ T} \)
  C) \( 1.91 \times 10^{-5} \text{ T} \)
  D) \( 3.82 \times 10^{-6} \text{ T} \)

Answer Key

• Q1: C
  Using \( B = \frac{\mu_0 I}{2 R} = \frac{4\pi \times 10^{-7} \times 2}{2 \times 0.05} = 1.26 \times 10^{-5} \text{ T} \)
• Q2: B
  Calculate \( B = \frac{\mu_0 I R^2}{2 (x^2 + R^2)^{3/2}} \) with given values gives \( 1.6 \times 10^{-6} \text{ T} \)
• Q3: A
  Use \( B = \frac{\mu_0 I \theta}{4 \pi R} \), \( \theta = \frac{\pi}{2} \), so \( B = \frac{4\pi \times 10^{-7} \times 3 \times \pi/2}{4 \pi \times 0.1} = 4.77 \times 10^{-6} \text{ T} \)

Memory Aids

• Magnetic field at center: \( B = \frac{\mu_0 I}{2 R} \)
• On axis at distance \( x \): \( B = \frac{\mu_0 I R^2}{2 (x^2 + R^2)^{3/2}} \)
• Arc field: \( B = \frac{\mu_0 I \theta}{4 \pi R} \)
• Right-hand thumb rule for direction
• For \( x \gg R \), \( B \propto \frac{1}{x^3} \) (dipole behavior)

Ampere's Circuital Law

Fundamentals

• Electric current \( I \) in a conductor produces magnetic field \( \mathbf{B} \) around it.
• Magnetic field lines form closed loops around the wire.
• Ampere's Law: \( \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 \sum I_{enc} \), where \( \mu_0 = 4\pi \times 10^{-7} \ \mathrm{T\cdot m/A} \).
• Right-hand rule: Thumb points in current direction; curled fingers show magnetic field direction.
• Only currents enclosed by the chosen loop contribute to the integral.

Limitations

• Applicable only for infinitely long, straight, and symmetrical current distributions.
• Requires known radius and constant current density for thick cylindrical wires.
• Current density \( J = \frac{I}{\pi R^2} \) must be uniform.

Solved Examples

Ex 1: Calculate magnetic field at distance \( r \) from a long straight wire carrying current \( I \).

• Use Ampere's law: \( B(2\pi r) = \mu_0 I \Rightarrow B = \frac{\mu_0 I}{2\pi r} \).

Practice Set

Q1. Magnetic field around a long straight wire carrying 5 A current at 0.1 m distance is:

• A) \( 1 \times 10^{-5} \ \mathrm{T} \)
• B) \( 1 \times 10^{-6} \ \mathrm{T} \)
• C) \( 1 \times 10^{-4} \ \mathrm{T} \)
• D) \( 1 \times 10^{-3} \ \mathrm{T} \)

Q2. For a loop enclosing currents \( I_1=3A \) out of page and \( I_2=2A \) into page, net current is:

• A) 5 A
• B) 1 A
• C) -1 A
• D) 0 A

Q3. Ampere's law is NOT applicable for:

• A) Infinite long wire
• B) Thick wire with uniform current density
• C) Finite irregular shaped conductor
• D) Symmetrical current distribution

Answer Key

Q1: A; \( B = \frac{4\pi \times 10^{-7} \times 5}{2\pi \times 0.1} = 1 \times 10^{-5} \ \mathrm{T} \).

Q2: B; Net current = \( 3 - 2 = 1 \) A (out of page positive).

Q3: C; Ampere's law fails for finite irregular conductors due to lack of symmetry.

Memory Aids

• Right-hand thumb = current; fingers = magnetic field.
• Magnetic field \( B \propto \frac{I}{r} \) outside wire.
• Ampere's law integral equals \( \mu_0 \times \) net enclosed current.

Magnetic Field in Solid Cylindrical Conductor

Key Concepts

• Conductor radius: \( R \), current: \( I \), current density: \( J = \frac{I}{\pi R^2} \) (constant).
• For \( r < R \) (inside wire): \( B_{in} = \frac{\mu_0 I r}{2\pi R^2} \) (increases linearly with \( r \)).
• At surface \( r = R \): \( B_s = \frac{\mu_0 I}{2\pi R} \) (maximum).
• For \( r > R \) (outside wire): \( B_{out} = \frac{\mu_0 I}{2\pi r} \) (decreases inversely with \( r \)).
• At center \( r=0 \), magnetic field \( B = 0 \).

Solved Examples

Ex 1: Find magnetic field at \( r = \frac{R}{2} \) inside the wire carrying current \( I \).

• Use \( B_{in} = \frac{\mu_0 I r}{2\pi R^2} = \frac{\mu_0 I (R/2)}{2\pi R^2} = \frac{\mu_0 I}{4\pi R} \).

Practice Set

Q1. Magnetic field at surface of wire with \( I=4A \), \( R=0.01m \) is:

• A) \( 4 \times 10^{-5} \ \mathrm{T} \)
• B) \( 8 \times 10^{-5} \ \mathrm{T} \)
• C) \( 2 \times 10^{-5} \ \mathrm{T} \)
• D) \( 1 \times 10^{-4} \ \mathrm{T} \)

Q2. Inside wire at \( r=0.005m \), magnetic field is:

• A) Half of surface field
• B) Equal to surface field
• C) Double surface field
• D) Zero

Q3. Magnetic field outside wire at \( r=0.02m \) is:

• A) \( \frac{\mu_0 I}{2\pi \times 0.02} \)
• B) \( \frac{\mu_0 I}{2\pi \times 0.01} \)
• C) Zero
• D) \( \frac{\mu_0 I}{4\pi \times 0.02} \)

Answer Key

Q1: A; \( B_s = \frac{4\pi \times 10^{-7} \times 4}{2\pi \times 0.01} = 4 \times 10^{-5} \ \mathrm{T} \).

Q2: A; \( B_{in} \propto r \), so at half radius, field is half surface field.

Q3: A; Outside field \( B_{out} = \frac{\mu_0 I}{2\pi r} \) with \( r=0.02m \).

Memory Aids

• Inside wire: \( B \propto r \) (linear increase).
• Surface: Maximum magnetic field.
• Outside wire: \( B \propto \frac{1}{r} \) (inverse relation).

The Solenoid

Definition and Structure

• Solenoid: A long coil of wire with length much greater than its radius.
• Used to produce a uniform magnetic field over a long range.
• Consists of conducting wire tightly wound in identical circular turns on a cylindrical frame.
• Magnetic field at any axial point is the superposition of fields from all circular turns.

Magnetic Field Inside Solenoid

• Magnetic field at an axial point inside a finite solenoid: \[ B = \frac{\mu_0 n I}{2} (\cos \theta_1 - \cos \theta_2) \] where \( n = \frac{N}{l} \) is the turn density, \( N \) total turns, \( l \) length, \( I \) current.
• For an infinite solenoid: \[ B = \mu_0 n I \] (uniform field inside)
• At the end of an infinite solenoid: \[ B = \frac{\mu_0 n I}{2} \]
• Magnetic field outside an infinite solenoid is approximately zero.
• If the solenoid frame has relative permeability \( \mu_r \), then: \[ B = \mu_0 \mu_r n I \]

This image shows a uniform magnetic field \( B \) directed along the axis of a straight conductor of length \( l \). Angles \( \theta_1 \) indicate the direction of the magnetic field at different points.

Graph of magnetic field strength \( B \) along the solenoid length: uniform and maximum at center \( \mu_0 n I \), decreasing to half at ends \( \frac{\mu_0 n I}{2} \).

Solved Examples

• Ex 1: Calculate magnetic field inside an infinite solenoid with 1000 turns/m and current 2 A.
  • Use \( B = \mu_0 n I \)
  • \( B = 4\pi \times 10^{-7} \times 1000 \times 2 = 2.51 \times 10^{-3} \text{ T} \)
• Ex 2: Find magnetic field at the end of the same solenoid.
  • Use \( B = \frac{\mu_0 n I}{2} \)
  • \( B = \frac{2.51 \times 10^{-3}}{2} = 1.25 \times 10^{-3} \text{ T} \)

Practice Set

• Q1: Magnetic field inside a solenoid depends on?
  • A) Number of turns per unit length and current
  • B) Radius of solenoid only
  • C) Length of solenoid only
  • D) Material outside solenoid
• Q2: Magnetic field at the end of an infinite solenoid is?
  • A) \( \mu_0 n I \)
  • B) \( \frac{\mu_0 n I}{2} \)
  • C) Zero
  • D) \( 2 \mu_0 n I \)
• Q3: A solenoid has 500 turns, length 0.5 m, current 3 A, and relative permeability 100. Calculate magnetic field inside.
  • A) \( 7.54 \times 10^{-3} \text{ T} \)
  • B) \( 1.51 \text{ T} \)
  • C) \( 0.15 \text{ T} \)
  • D) \( 0.75 \text{ T} \)

Answer Key

• Q1: A — Magnetic field depends on turn density \( n \) and current \( I \).
• Q2: B — At the end, field is half the inside value \( \frac{\mu_0 n I}{2} \).
• Q3: B — Calculate \( n = \frac{500}{0.5} = 1000 \), then \( B = \mu_0 \mu_r n I = 4\pi \times 10^{-7} \times 100 \times 1000 \times 3 = 1.51 \text{ T} \).

Memory Aids

• Inside infinite solenoid: \( B = \mu_0 n I \)
• At ends: field halves \( \Rightarrow B = \frac{\mu_0 n I}{2} \)
• Field outside infinite solenoid ≈ 0
• Use \( n = \frac{N}{l} \) for turn density
• Include \( \mu_r \) if magnetic material present: \( B = \mu_0 \mu_r n I \)

Force between Two Parallel Currents, the Ampere

Magnetic Force Between Wires

• Two parallel wires carry currents \(I_1\) and \(I_2\) separated by distance \(d\).
• Each wire produces magnetic field: \(B = \frac{\mu_0 I}{2 \pi d}\), where \(\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}\).
• Force on length \(dl\) of wire 1 due to wire 2: \(dF_{12} = B_2 I_1 dl = \frac{\mu_0 I_1 I_2}{2 \pi d} dl\).
• Force per unit length on each wire: \(\frac{dF}{dl} = \frac{\mu_0 I_1 I_2}{2 \pi d}\).
• Direction: Currents in same direction attract; opposite directions repel.
• Definition of ampere: Two wires 1 m apart carrying 1 A each experience force \(2 \times 10^{-7} \text{ N/m}\).

Equilibrium of Free Wire

• Consider two wires: one fixed, one free, separated by height \(h\).
• Mass per unit length of free wire: \(\lambda = \frac{m}{l}\).
• Weight per unit length: \(W = \lambda g\).
• Stable equilibrium condition: \(\frac{\mu_0 I_1 I_2}{2 \pi h} = \lambda g\).
• If free wire displaced slightly, it executes vertical simple harmonic motion (S.H.M.).
• Time period of S.H.M.: \(T = 2 \pi \sqrt{\frac{h}{q}}\), where \(q\) depends on system parameters.
• Unstable equilibrium if lower wire is free: small displacement causes wire to move away.

Solved Examples

Ex 1

• Two parallel wires 0.5 m apart carry currents 3 A and 4 A in same direction.
• Find force per unit length between them.
• Use formula: \(\frac{dF}{dl} = \frac{\mu_0 I_1 I_2}{2 \pi d} = \frac{4\pi \times 10^{-7} \times 3 \times 4}{2 \pi \times 0.5} = 4.8 \times 10^{-6} \text{ N/m}\).

Ex 2

• Free wire of mass per unit length \(\lambda = 0.01 \text{ kg/m}\) is suspended at height 0.2 m from fixed wire carrying current 5 A.
• Find current in free wire for stable equilibrium.
• Use \(\frac{\mu_0 I_1 I_2}{2 \pi h} = \lambda g\), solve for \(I_2\): \(I_2 = \frac{2 \pi h \lambda g}{\mu_0 I_1} = \frac{2 \pi \times 0.2 \times 0.01 \times 9.8}{4\pi \times 10^{-7} \times 5} = 3920 \text{ A}\).

Practice Set

Q1

Two parallel wires 1 m apart carry currents 2 A and 3 A in opposite directions. What is the force per meter between them?

• A) \(1.2 \times 10^{-6} \text{ N}\)
• B) \(3.6 \times 10^{-7} \text{ N}\)
• C) \(1.2 \times 10^{-7} \text{ N}\)
• D) \(3.6 \times 10^{-6} \text{ N}\)

Q2

Two wires 0.25 m apart carry currents 5 A and 10 A in same direction. Calculate force per unit length.

• A) \(4 \times 10^{-5} \text{ N/m}\)
• B) \(2 \times 10^{-5} \text{ N/m}\)
• C) \(8 \times 10^{-6} \text{ N/m}\)
• D) \(1 \times 10^{-5} \text{ N/m}\)

Q3

A free wire of length 0.5 m and mass 0.02 kg is suspended at height 0.1 m from a fixed wire carrying 4 A. Find current in free wire for stable equilibrium.

• A) 490 A
• B) 980 A
• C) 245 A
• D) 1960 A

Answer Key

• Q1: A) \(1.2 \times 10^{-6} \text{ N}\) - Use \(\frac{\mu_0 I_1 I_2}{2 \pi d}\) with \(d=1\) m.
• Q2: A) \(4 \times 10^{-5} \text{ N/m}\) - Calculate force per unit length with given currents and distance.
• Q3: D) 1960 A - Use stable equilibrium formula \(I_2 = \frac{2 \pi h \lambda g}{\mu_0 I_1}\) with \(\lambda = \frac{m}{l}\).

Memory Aids

• Force per unit length \(\propto I_1 I_2 / d\).
• Same current direction → attraction; opposite → repulsion.
• Ampere defined by force \(2 \times 10^{-7} \text{ N/m}\) at 1 m distance.
• Stable equilibrium: magnetic force balances weight \(\Rightarrow \frac{\mu_0 I_1 I_2}{2 \pi h} = \lambda g\).
• Displaced free wire executes S.H.M. with period \(T = 2 \pi \sqrt{\frac{h}{q}}\).

Torque on Current Loop

Key Concepts

• Rectangular loop PQRS with sides a and b carries current I.
• Placed in uniform magnetic field \( \vec{B} \), area vector \( \vec{A} \) makes angle \( \theta \) with \( \vec{B} \).
• Forces on sides QR and SP are equal, opposite, and collinear; they cancel out.
• Forces on sides PQ and RS are equal, opposite, but not collinear; produce torque.
• Force on PQ or RS: \( F = I b B \sin 90^\circ = I b B \).
• Perpendicular distance between forces: \( r_\perp = a \sin \theta \).
• Torque on loop: \( \tau = F r_\perp = I b B a \sin \theta = I A B \sin \theta \), where \( A = a b \).
• For loop with N turns: \( \tau = N I A B \sin \theta \).
• Vector form: \( \vec{\tau} = \vec{M} \times \vec{B} \), where magnetic dipole moment \( \vec{M} = N I \vec{A} \).
• No torque when \( \theta = 0^\circ \) (loop plane normal to \( \vec{B} \)); torque maximum at \( \theta = 90^\circ \).

Solved Examples

• Ex 1: Calculate torque on a rectangular loop with \( I=2\,A \), \( a=0.1\,m \), \( b=0.2\,m \), \( B=0.5\,T \), and \( \theta=30^\circ \).
  • Area \( A = 0.1 \times 0.2 = 0.02\,m^2 \).
  • Torque \( \tau = I A B \sin \theta = 2 \times 0.02 \times 0.5 \times \sin 30^\circ = 0.01\,Nm \).

Practice Set

• Q1: A rectangular loop with area \( 0.05\,m^2 \) carries current \( 3\,A \) in a magnetic field of \( 0.4\,T \). If \( \theta = 90^\circ \), what is the torque?
  • A) 0.06 Nm
  • B) 0.12 Nm
  • C) 0.15 Nm
  • D) 0.20 Nm
• Q2: For a loop with \( N=5 \) turns, \( I=1\,A \), \( A=0.01\,m^2 \), \( B=0.3\,T \), and \( \theta=60^\circ \), torque is:
  • A) 0.0075 Nm
  • B) 0.015 Nm
  • C) 0.025 Nm
  • D) 0.03 Nm
• Q3: Torque on a single-turn loop is \( 0.02\,Nm \) in a field \( 0.5\,T \) with current \( 2\,A \). Find the area if \( \theta=45^\circ \).
  • A) 0.0283 m²
  • B) 0.0141 m²
  • C) 0.0566 m²
  • D) 0.0707 m²

Answer Key

• Q1: A) 0.06 Nm — \( \tau = IAB\sin 90^\circ = 3 \times 0.05 \times 0.4 = 0.06 \) Nm.
• Q2: B) 0.015 Nm — \( \tau = N I A B \sin 60^\circ = 5 \times 1 \times 0.01 \times 0.3 \times 0.866 = 0.013 \approx 0.015 \) Nm.
• Q3: B) 0.0141 m² — \( A = \frac{\tau}{I B \sin \theta} = \frac{0.02}{2 \times 0.5 \times 0.707} = 0.0141 \) m².

Memory Aids

• Torque \( \tau = N I A B \sin \theta \).
• Magnetic dipole moment \( \vec{M} = N I \vec{A} \).
• Torque vector \( \vec{\tau} = \vec{M} \times \vec{B} \).
• No torque when loop plane is perpendicular to \( \vec{B} \) (\( \theta=0^\circ \)).

Circular Current Loop Magnetic Dipole

Key Concepts

• Current loop of radius \( r \) carries current \( I \).
• Magnetic field at distance \( x \) on axis of loop: \[ B = \frac{\mu_0 I r^2}{2 (r^2 + x^2)^{3/2}} = \frac{\mu_0 I A}{2 \pi (r^2 + x^2)^{3/2}} \]
• Area of loop \( A = \pi r^2 \).
• For \( x \gg r \), field approximates to: \[ B = \frac{\mu_0 I A}{2 \pi x^3} = \frac{\mu_0}{4 \pi} \frac{2 M}{x^3} \]
• Magnetic dipole moment \( M = I A \).

Solved Examples

• Ex 1: Find magnetic field at \( x=0.1\,m \) on axis of loop with \( I=3\,A \), \( r=0.05\,m \).
  • Area \( A = \pi (0.05)^2 = 7.85 \times 10^{-3} m^2 \).
  • Calculate \( B = \frac{4\pi \times 10^{-7} \times 3 \times 7.85 \times 10^{-3}}{2 \pi (0.05^2 + 0.1^2)^{3/2}} \approx 1.2 \times 10^{-5} T \).

Practice Set

• Q1: Magnetic field at center (\( x=0 \)) of loop with \( I=2\,A \), \( r=0.1\,m \) is:
  • A) \( 2 \times 10^{-6} T \)
  • B) \( 4 \times 10^{-6} T \)
  • C) \( 1.26 \times 10^{-5} T \)
  • D) \( 5 \times 10^{-6} T \)
• Q2: For \( x \gg r \), magnetic field varies as:
  • A) \( 1/x \)
  • B) \( 1/x^2 \)
  • C) \( 1/x^3 \)
  • D) \( 1/x^4 \)
• Q3: Magnetic dipole moment of loop with \( I=4\,A \), \( r=0.05\,m \) is:
  • A) \( 3.14 \times 10^{-2} A m^2 \)
  • B) \( 1.57 \times 10^{-2} A m^2 \)
  • C) \( 6.28 \times 10^{-3} A m^2 \)
  • D) \( 7.85 \times 10^{-3} A m^2 \)

Answer Key

• Q1: C) \( 1.26 \times 10^{-5} T \) — At center, \( B = \frac{\mu_0 I}{2r} = 1.26 \times 10^{-5} T \).
• Q2: C) \( 1/x^3 \) — Field on axis far from loop decreases as inverse cube of distance.
• Q3: D) \( 7.85 \times 10^{-3} A m^2 \) — \( M = I A = 4 \times \pi (0.05)^2 = 7.85 \times 10^{-3} A m^2 \).

Memory Aids

• Magnetic field on axis: \( B = \frac{\mu_0 I A}{2 \pi (r^2 + x^2)^{3/2}} \).
• Far field: \( B \propto \frac{1}{x^3} \).
• Magnetic dipole moment: \( M = I A \).

The Moving Coil Galvanometer

Principle and Working

• A current-carrying coil in a magnetic field experiences a torque \( \tau = N I \mathbf{A} \times \mathbf{B} \).
• Here, \(N\) = number of turns, \(I\) = current, \(\mathbf{A}\) = area vector, \(\mathbf{B}\) = magnetic field.
• Torque causes coil deflection \( \Phi \), opposed by spring's restoring torque \( k \Phi \).
• At equilibrium: \( N I A B = k \Phi \) or \( I = \frac{k}{N A B} \Phi \).
• Current \(I\) is proportional to deflection \(\Phi\); constant \( \frac{k}{NAB} \) is galvanometer constant.
• Cylindrical pole pieces ensure uniform radial magnetic field.

Current and Voltage Sensitivity

• Current sensitivity \( = \frac{\Phi}{I} = \frac{NAB}{k} \) (deflection per unit current).
• Increasing \(N\) increases current sensitivity.
• Voltage sensitivity \( = \frac{\Phi}{V} = \frac{\Phi}{I} \times \frac{I}{V} = \frac{NAB}{k} \times \frac{1}{R} \) (deflection per unit voltage).
• Doubling \(N\) doubles current sensitivity but also doubles resistance \(R\), so voltage sensitivity remains unchanged.

Conversion to Ammeter

• Connect a low resistance shunt \( r_s \) in parallel with galvanometer.
• Shunt carries most current, protecting galvanometer.
• Shunt resistance: \( r_s = \frac{I_g}{I - I_g} R_G \), where \(I_g\) is galvanometer current, \(I\) total current, \(R_G\) galvanometer resistance.

Conversion to Voltmeter

• Connect a large resistance \( R \) in series with galvanometer.
• Limits current through galvanometer for measuring voltage.
• Resistance \( R = \frac{V}{I_g} - R_G \), where \(V\) is voltage, \(I_g\) galvanometer current, \(R_G\) galvanometer resistance.

Solved Examples

Ex 1

• Given \( N=100, A=2 \times 10^{-4} \text{ m}^2, B=0.5 \text{ T}, k=0.02 \text{ Nm/rad} \), find current for \( \Phi=0.1 \text{ rad} \).
• Use \( I = \frac{k}{NAB} \Phi = \frac{0.02}{100 \times 2 \times 10^{-4} \times 0.5} \times 0.1 = 0.02 \text{ A} \).

Ex 2

• Galvanometer resistance \( R_G=50 \Omega \), current sensitivity \( \frac{\Phi}{I} = 100 \text{ rad/A} \), find shunt resistance to convert to ammeter for max current \( I=2 \text{ A} \) and max galvanometer current \( I_g=0.01 \text{ A} \).
• Use \( r_s = \frac{I_g}{I - I_g} R_G = \frac{0.01}{2 - 0.01} \times 50 \approx 0.25 \Omega \).

Practice Set

Q1

What is the relation between current \(I\) and deflection \(\Phi\) in a moving coil galvanometer?

• A) \( I \propto \Phi^2 \)
• B) \( I \propto \Phi \)
• C) \( I \propto \frac{1}{\Phi} \)
• D) \( I \propto \sqrt{\Phi} \)

Q2

Doubling the number of turns \(N\) in the coil affects the current sensitivity and voltage sensitivity as:

• A) Both double
• B) Current sensitivity doubles, voltage sensitivity unchanged
• C) Current sensitivity unchanged, voltage sensitivity doubles
• D) Both unchanged

Q3

A galvanometer with resistance \(50 \Omega\) is converted into an ammeter to measure 5 A max current. If max galvanometer current is 0.05 A, find the shunt resistance \(r_s\).

• A) 0.5 \(\Omega\)
• B) 0.51 \(\Omega\)
• C) 1 \(\Omega\)
• D) 5 \(\Omega\)

Answer Key

• Q1: B) \( I \propto \Phi \). Current is directly proportional to deflection.
• Q2: B) Current sensitivity doubles, voltage sensitivity unchanged. Doubling \(N\) doubles current sensitivity but also doubles resistance, keeping voltage sensitivity same.
• Q3: B) \( r_s = \frac{I_g}{I - I_g} R_G = \frac{0.05}{5 - 0.05} \times 50 = 0.51 \Omega \).

Memory Aids

• Current \(I\) ∝ deflection \(\Phi\): \( I = \frac{k}{NAB} \Phi \).
• Current sensitivity \(= \frac{NAB}{k} \), increase \(N\) to increase sensitivity.
• Voltage sensitivity \(= \frac{NAB}{kR} \), depends inversely on resistance.
• Shunt resistance for ammeter: \( r_s = \frac{I_g}{I - I_g} R_G \).
• Series resistance for voltmeter: \( R = \frac{V}{I_g} - R_G \).