Comprehensive Study Notes: Moving Charges and Magnetism

INTRODUCTION

  • Historical Context: Historically, it was believed that electricity and magnetism were independent phenomena with no connection between them.

  • Christian Oersted’s Observation (1820): Hans Christian Oersted discovered that when a compass needle is placed very close to a current-carrying conductor, the needle deflects. This led to the conclusion that just as static charges produce an electric field, a current (or moving charges) produces a magnetic field in addition to any existing electric field.

MAGNETIC FIELD AND INDUCTION

  • Definition of Magnetic Field: It is the region surrounding a magnetic dipole or a current-carrying conductor within which other magnetic dipoles or current-carrying conductors experience a force of attraction or repulsion.

  • Magnetic Induction Vector (BB): The strength and direction of a magnetic field are defined by a vector quantity called magnetic induction. It is also referred to as:

    • Magnetic Flux Density

    • Magnetic Field (BB)

  • Units of Magnetic Field:

    • weber/metre2\text{weber/metre}^2

    • tesla (T)\text{tesla (T)}

  • Magnetic Field Lines: These are imaginary lines within a magnetic field representing the path along which a hypothetical, isolated north pole would move.

  • Magnetic Flux (ΦB\Phi_B): This represents the total number of magnetic field lines passing through a specific unit area in a direction perpendicular to that area. It is defined mathematically as:     ΦB=∫B⋅da\Phi_B = \int B \cdot da

MAGNETIC LORENTZ FORCE

  • Definition: When a charge qq moves with a velocity VV inside a magnetic field BB, it experiences a magnetic force (FBF_B) known as the Magnetic Lorentz Force:     FB=q(V×B)F_B = q(V \times B)

  • Key Characteristics of Magnetic Lorentz Force:

    1. The force depends on the magnitude of the charge qq, the velocity VV, and the magnetic field BB. Its direction is perpendicular to both VV and BB, determined by the right-hand thumb rule.

    2. The force acting on a negative charge is in the exact opposite direction of the force acting on a positive charge.

    3. F=0F = 0 if the velocity V=0V = 0 (stationary charge) or if the magnetic field B=0B = 0.

    4. F=0F = 0 if the velocity VV and magnetic field BB are parallel (θ=0∘\theta = 0^{\circ}) or anti-parallel (θ=180∘\theta = 180^{\circ}).

  • Total Lorentz Force: This is the total force acting on a charged particle moving through a region where both electric (EE) and magnetic (BB) fields coexist:     F=FE+FB=qE+q(V×B)F = F_E + F_B = qE + q(V \times B)     F=q(E+V×B)F = q(E + V \times B)

WORK DONE BY MAGNETIC FORCE

  • Principles of Work: Given the magnetic force FB=q(V×B)F_B = q(V \times B), the work done (dWdW) by this force on a charged particle is calculated as:     dW=FB⋅drdW = F_B \cdot dr

  • Conclusion: Since the displacement drdr and velocity VV are in the same direction and both are always perpendicular to the magnetic force FBF_B, the dot product is zero:     dW=FB⋅dr=0dW = F_B \cdot dr = 0     Therefore, the magnetic force does no work on the charged particle.

MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD

  • Case 1: Movement Parallel to the Field (V∥BV \parallel B)

    • θ=0∘\theta = 0^{\circ} or 180∘180^{\circ}

    • FB=qVBsin⁡(0)=0F_B = qVB \sin(0) = 0

    • The particle experiences no force and continues to move in a straight line, unaffected by the field.

  • Case 2: Movement Perpendicular to the Field (V⊥BV \perp B)

    • θ=90∘\theta = 90^{\circ}

    • F=qVBsin⁡(90∘)=qVBF = qVB \sin(90^{\circ}) = qVB

    • This force acts perpendicular to the velocity at every point, serving as a centripetal force. Consequently, the particle executes circular motion perpendicular to the magnetic field.

    • Derivation of Radius (rr):         mv2r=qVB  ⟹  r=mVqB\frac{mv^2}{r} = qVB \implies r = \frac{mV}{qB}

    • Time Period (TT):         T=2πrV=2πmqBT = \frac{2\pi r}{V} = \frac{2\pi m}{qB}

    • Frequency of Rotation (ν\nu):         ν=1T=qB2πm\nu = \frac{1}{T} = \frac{qB}{2\pi m}

    • Observation: Frequency is independent of velocity. This principle is utilized in the design of the cyclotron. As momentum (mVmV) increases, the radius (rr) also increases.

  • Case 3: Entry at an Arbitrary Angle (θ\theta)

    • The velocity has two components:

      • Perpendicular Component (V⊥=Vsin⁡(θ)V_{\perp} = V \sin(\theta)): Rotates the particle.

      • Parallel Component (V∥=Vcos⁡(θ)V_{\parallel} = V \cos(\theta)): Translates the particle along the field lines.

    • Result: The particle executes helical motion.

    • Pitch (PP): The distance moved along the magnetic field during one full rotation.         P=T×V∥P = T \times V_{\parallel}         P=2πmVcos⁡(θ)qBP = \frac{2\pi m V \cos(\theta)}{qB}

NUMERICAL EXAMPLE: ELECTRON PATH

  • Problem: Calculate the radius of the path, frequency, and energy (in keV) for an electron (m=9×10−31 kgm = 9 \times 10^{-31}\,kg, q=1.6×10−19 Cq = 1.6 \times 10^{-19}\,C) moving at a speed of 3×107 m/s3 \times 10^7\,m/s in a magnetic field of 6×10−4 T6 \times 10^{-4}\,T perpendicular to it.

  • Radius Calculation:     r=mVqB=9×10−31×3×1071.6×10−19×6×10−4=26×10−2 mr = \frac{mV}{qB} = \frac{9 \times 10^{-31} \times 3 \times 10^7}{1.6 \times 10^{-19} \times 6 \times 10^{-4}} = 26 \times 10^{-2}\,m

  • Frequency Calculation:     ν=qB2πm=2×107 Hz\nu = \frac{qB}{2\pi m} = 2 \times 10^7\,Hz

  • Energy Calculation:     E=12mV2=40.5×10−17 JE = \frac{1}{2}mV^2 = 40.5 \times 10^{-17}\,J     Conversion: 1 eV=1.6×10−19 J1\,eV = 1.6 \times 10^{-19}\,J     E=2.5 keVE = 2.5\,keV

FORCE ON A CURRENT-CARRYING CONDUCTOR

  • Scenario: A wire with cross-section AA, length ll, free electron density nn, carrying current II in an external field BB.

  • Derivation:

    • Total charge in the wire: q=nAleq = nAle

    • Average electron velocity: Drift velocity (VdV_d)

    • Total force (FBF_B): nAle(Vd×B)nAle(V_d \times B)

    • Since I=neAVdI = neAV_d, the equation simplifies for a straight conductor to:         FB=I(l×B)F_B = I(l \times B)

  • Arbitrary Shapes: For a wire of non-linear shape, the force on a small element is:     FB=I(dl×B)F_B = I(dl \times B)

BIOT-SAVART LAW

  • Definition: The magnetic field (dBdB) at a point due to a small current element (IdlIdl) is:

    1. Directly proportional to current (II).

    2. Directly proportional to the length of the element (dldl).

    3. Directly proportional to the sine of the angle (θ\theta) between the element and the displacement vector rr.

    4. Inversely proportional to the square of the distance (r2r^2).

  • Mathematical Formula:     dB=μ04πIdlsin⁡(θ)r2dB = \frac{\mu_0}{4\pi} \frac{I dl \sin(\theta)}{r^2}     Where μ04π=10−7 Tm/A\frac{\mu_0}{4\pi} = 10^{-7}\,Tm/A is the proportionality constant.

  • Vector Form:     dB=μ04πI(dl×r)r3dB = \frac{\mu_0}{4\pi} \frac{I(dl \times r)}{r^3}

MAGNETIC FIELD ON THE AXIS OF A CIRCULAR LOOP

  • Setup: Loop of radius RR in the YZ plane, centered at the origin, carrying current II. Axis is the X-axis.

  • Calculation at distance xx from center:

    1. The magnetic field dBdB from an element has X-components (dBcos⁡(θ)dB\cos(\theta)) and perpendicular components (dBsin⁡(θ)dB\sin(\theta)).

    2. Perpendicular components cancel out when summed over the loop.

    3. Net field B=∫dBcos⁡(θ)B = \int dB\cos(\theta).

    4. Resulting formula:         B=μ0IR22(x2+R2)3/2B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}

  • Special Cases:

    1. For NN turns: B=μ0NIR22(x2+R2)3/2B = \frac{\mu_0 N I R^2}{2(x^2 + R^2)^{3/2}}

    2. At the center of the loop (x=0x=0): B=μ0I2RB = \frac{\mu_0 I}{2R}

AMPERE’S CIRCUITAL LAW

  • Definition: The line integral of the magnetic field around any closed loop is equal to μ0\mu_0 times the net current (IencI_{enc}) enclosed by that loop:     ∮B⋅dl=μ0Ienc\oint B \cdot dl = \mu_0 I_{enc}

  • Application: Long Straight Conductor:

    • For a point at distance rr, ∮B⋅dl=B(2πr)\oint B \cdot dl = B(2\pi r).

    • By Ampere's Law: B(2πr)=μ0IB(2\pi r) = \mu_0 I

    • Formula: B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

MAGNETIC FIELD OF A SOLENOID

  • Definition: A solenoid is a long insulated wire wound in a helix where turns are closely spaced. It behaves like a bar magnet when current flows.

  • Ideal Solenoid: Field outside is nearly zero; field inside is parallel to the axis and uniform.

  • Magnetic Field Formula:     B=μ0nIB = \mu_0 n I     Where n=NLn = \frac{N}{L} (turns per unit length).

  • Ways to Increase Magnetic Field (BB):

    1. Increase the number of turns per unit length.

    2. Increase the current flow.

    3. Insert a soft iron core into the solenoid.

FORCE BETWEEN PARALLEL CONDUCTORS (AMPERE FORCE)

  • Scenario: Two straight conductors PP and QQ separated by distance rr carrying currents I1I_1 and I2I_2.

  • Force per unit length (ff):     f=μ0I1I22πrf = \frac{\mu_0 I_1 I_2}{2\pi r}

  • Interaction Rules:

    • Currents in the same direction: Forces are attractive.

    • Currents in opposite directions: Forces are repulsive.

  • Definition of One Ampere: An ampere is the constant current which, if maintained in two straight parallel conductors of infinite length placed one metre apart in vacuum, produces a force per unit length of 2×10−7 N/m2 \times 10^{-7}\,N/m between them.

TORQUE ON A CURRENT LOOP AND MAGNETIC DIPOLE

  • Torque (τ\tau) in Uniform Magnetic Field:     τ=m×B=mBsin⁡(θ)\tau = m \times B = m B \sin(\theta)     For a rectangular loop with turns NN, current II, and area AA:     τ=NIABsin⁡(θ)\tau = NIAB \sin(\theta)

  • Magnetic Dipole Moment (mm):     m=IAm = IA     For NN turns: m=NIAm = NIA

  • Comparison to Electric Dipole:

    • Magnetic Field at axial distance xx: B=μ04π2mx3B = \frac{\mu_0}{4\pi} \frac{2m}{x^3}

    • Electric Field at axial distance xx: E=14πϵ02px3E = \frac{1}{4\pi\epsilon_0} \frac{2p}{x^3}

    • This confirms a circular loop acts as a magnetic dipole.

MOVING COIL GALVANOMETER

  • Purpose: To detect the presence and direction of small electric currents.

  • Construction: Light rectangular coil with NN turns and area AA wound on a soft iron cylinder. It rotates in a uniform radial magnetic field created by concave magnetic poles. A spring provides a restoring torque.

  • Working Principle: Torque produced by current (NIABNIAB) is balanced by restoring torque from the spring (kϕk\phi):     NIAB=kϕ  ⟹  I=(kNAB)ϕNIAB = k\phi \implies I = (\frac{k}{NAB}) \phi     Thus, current is directly proportional to deflection (I∝ϕI \propto \phi).

  • Sensitivities:

    • Current Sensitivity: Deflection per unit current = ϕI=NABk\frac{\phi}{I} = \frac{NAB}{k}

    • Voltage Sensitivity: Deflection per unit voltage = ϕV=NABkR\frac{\phi}{V} = \frac{NAB}{kR}

  • Note: Doubling the number of turns (NN) doubles current sensitivity but might not change voltage sensitivity because doubling NN often doubles resistance (RR).

AMMETER AND VOLTMETER CONVERSION

  • Ammeter (Current Measurement):

    • Purpose: High sensitivity current detection must be modified to measure large currents without damage.

    • Conversion: Connect a small resistance (shunt SS) in parallel with the galvanometer (GG).

    • Formula: S=IgGI−IgS = \frac{I_g G}{I - I_g}

    • Note: Ideal ammeter resistance is zero. It is always connected in series.

  • Voltmeter (Potential Difference Measurement):

    • Purpose: Measure potential difference between two points without affecting current flow.

    • Conversion: Connect a high resistance (RR) in series with the galvanometer (GG).

    • Formula: R=VIg−GR = \frac{V}{I_g} - G

    • Note: Ideal voltmeter resistance is infinite. It is always connected in parallel.

QUESTIONS & DISCUSSION

  • Question (March 2020): Why might increasing the current sensitivity not increase the voltage sensitivity?

    • Response: If resistance increases proportionally with the number of turns (e.g., doubling NN), the ratio NR\frac{N}{R} remains constant, hence voltage sensitivity remains unchanged.

  • Question (March 2024): Why is an ammeter always connected in series?

    • Response: To ensure that the entire circuit current passes through the instrument for measurement while maintaining a low resistance so as not to change the circuit's total resistance.

  • Question (March 2025): What will be the direction of magnetic field with respect to the velocity of a charged particle moving in a circular path?

    • Response: The magnetic field must be perpendicular (90∘90^{\circ}) to the velocity of the charged particle to sustain circular motion.

  • Question (March 2025): Is there any change in kinetic energy of the charged particle in a magnetic field?

    • Response: No, because the magnetic force is always perpendicular to the velocity, meaning it does no work. Therefore, the speed (and kinetic energy) remains constant, even though the direction of motion changes.