Magnetic Effects of Current – Detailed Examination Notes

Page 1 – Historical Background & Fundamental Definitions

  • 1820: Hans Christian Øersted discovers that electric current in a straight conductor deflects a compass needle.
    • Deflection ∝ magnitude of current.
    • Reversing current reverses deflection.
  • Ampère’s “Swimming Rule” (Mnemonic – SNOW)
    1. Imagine a man swimming along the wire in the direction of current; current enters his feet, leaves his head.
    2. With his face toward the needle, the north pole of the compass is deflected toward his left hand.
    3. SNOW: Current from South to North over a magnetic needle deflects its North pole toward West.
  • Magnetic field (magnetic induction, magnetic flux density) B
    • Region around a current-carrying conductor or magnet where magnetic effects are felt.
    • Disappears when current stops (moving charge needed).
    • Moving charge acts as a source of both electric and magnetic fields.
    • B\vec B is a vector.
  • Biot–Savart Law – qualitative statement
    • Deals with magnetic induction at a point due to elemental current segment IdlI\,d\vec l.
    • Geometry: point P at distance rr from element; angle between dld\vec l and r\vec r is θ\theta.

Page 2 – Biot–Savart Law: Mathematical Form & Consequences

  • Proportionalities for elemental field dBd\vec B:
    1. dBIdB \propto I
    2. dBdldB \propto dl
    3. dBsinθdB \propto \sin \theta
    4. dB1r2dB \propto \frac{1}{r^{2}}
  • Combining and inserting the constant μ<em>0/4π\mu<em>0/4\pi (SI) or k</em>mk</em>m (CGS):
    dB=μ04πIdl×r^r2d\vec B = \frac{\mu_0}{4\pi}\, \frac{I\, d\vec l \times \hat r}{r^{2}}
    where r^\hat r is the unit vector from element to P.
  • μ0\mu_0 (absolute permeability of free space) = 4π×107  WbA1m14\pi \times 10^{-7}\;\mathrm{Wb\,A^{-1}m^{-1}}.
  • Direction: right-hand screw rule (dld\vec l → tip of screw, rotation gives B\vec B).
  • For entire current path: integrate
    B=μ04πIdl×r^r2\vec B = \frac{\mu_0}{4\pi} \int \frac{I\, d\vec l \times \hat r}{r^{2}}
  • Current density form (use JdV=Idl\vec J\, dV = I d\vec l):
    B=μ04πJ×r^r2dV\vec B = \frac{\mu_0}{4\pi} \int \frac{\vec J \times \hat r}{r^{2}}\, dV
  • Comparisons with Coulomb’s law:
    1. Both long-range ( 1/r2\propto 1/r^{2} ).
    2. Superposition principle holds.
    3. Electrostatic source is scalar charge qq; magnetostatic source is vector IdlI d\vec l.
    4. Electrostatic field along r\vec r; magnetic field perpendicular to plane of IdlI d\vec l and r\vec r.
    5. Angle factor sinθ\sin\theta present only in Biot–Savart.

Page 3 – Relation among μ<em>0\mu<em>0, ε</em>0\varepsilon</em>0, and cc

  • 14πε0=9×109  Nm2C2\frac{1}{4\pi\varepsilon_0}=9\times10^{9}\;\mathrm{N\,m^{2}C^{-2}}
  • μ0=4π×107  WbA1m1\mu_0 = 4\pi \times 10^{-7}\;\mathrm{Wb\,A^{-1}m^{-1}}
  • Using Maxwell’s theory: μ<em>0ε</em>0=1c2\mu<em>0\varepsilon</em>0 = \frac{1}{c^{2}}
    c=1μ<em>0ε</em>03×108  ms1c = \frac{1}{\sqrt{\mu<em>0\varepsilon</em>0}} \approx 3\times10^{8}\;\mathrm{m\,s^{-1}}

Page 3 (cont.) – Magnetic Field on Axis of a Circular Coil

  • Coil radius aa, current II, point P on axis at distance xx from center.
  • From two diametrically opposite elements dld\vec l, transverse components cancel; axial components add.
  • Result for single turn:
    B=μ0Ia22(a2+x2)3/2B = \frac{\mu_0 I a^{2}}{2\,(a^{2}+x^{2})^{3/2}}
  • For N turns:
    B=μ0NIa22(a2+x2)3/2B = \frac{\mu_0 N I a^{2}}{2\,(a^{2}+x^{2})^{3/2}}
  • Special cases
    1. At centre (x=0x=0): B=μ0NI2aB = \frac{\mu_0 N I}{2a}
    2. Far-field (xax \gg a): Bμ0NIa22x3B \approx \frac{\mu_0 N I a^{2}}{2x^{3}} (dipole behaviour).

Page 5 – Direction Rules & Variation Along Axis

  • Right-hand thumb rule: curl fingers along current, thumb gives B\vec B.
  • “Clock rule” for faces of a coil
    • Anticlockwise current → north pole (outward B\vec B).
    • Clockwise current → south pole.
  • Variation of BB along axis (graphically decreases from centre to zero at infinity as above formula).

Page 5 (end) & Page 6 – Centre of Loop & Ampère’s Circuital Law

  • For loop of radius RR, centre field:
    B=μ0NI2RB = \frac{\mu_0 N I}{2R}
  • Ampère’s Circuital Law: for closed path Bdl=μ<em>0I</em>enc\oint \vec B \cdot d\vec l = \mu<em>0 I</em>{\text{enc}}.
    • Proof sketched with infinitely long straight conductor: choose circular Amperian loop radius rr.
    • Since BB constant on loop and tangential: B(2πr)=μ<em>0IB=μ</em>0I2πrB(2\pi r)=\mu<em>0 I \Rightarrow B=\frac{\mu</em>0 I}{2\pi r}.

Page 7 – Infinite Straight Wire & Solenoid

  • Infinite straight wire: B=μ0I/(2πr)B=\mu_0 I / (2\pi r) (direction via right-hand rule around wire).
  • Solenoid (length ll\gg diameter, N total turns): turn density n=N/ln=N/l.
    • Inside (near axis): uniform B=μ<em>0nI=μ</em>0NI/lB=\mu<em>0 n I = \mu</em>0 N I / l.
    • At ends ≈ 12μ0nI\tfrac{1}{2} \mu_0 n I.
  • Finite solenoid: outside field weak because contributions cancel; inside strong & uniform.

Page 8 – Toroid

  • Solenoid bent into closed ring (toroid), N turns, current I.
  • Inside core (radius r between inner & outer boundaries): B=μ0NI/(2πr)B=\mu_0 N I /(2\pi r) (non-uniform, inversely with r).
  • Outside toroid: B=0B=0 using Ampère’s law (net current enclosed =0).

Page 9 – Magnetic Force on a Moving Charge

  • Lorentz magnetic force: F=qv×B\vec F = q\, \vec v \times \vec B
    • Magnitude F=qvBsinθF = q v B \sin\theta.
    • θ\theta = angle between v\vec v and B\vec B.
    • Direction from right-hand-screw or Fleming’s left-hand rule (for conventional current).
  • Special cases
    1. θ=0,π\theta = 0,\,\piF=0F=0 (motion parallel/antiparallel to B\vec B).
    2. θ=90\theta = 90^{\circ}Fmax=qvBF_{\max}=qvB.
  • SI unit of BB (Tesla): 1T=1NA1m1=1Wbm21\,\mathrm{T}=1\,\mathrm{N\,A^{-1}m^{-1}}=1\,\mathrm{Wb\,m^{-2}}. Defined such that q=1C,v=1m/s,F=1N,θ=90q=1\,\mathrm{C}, v=1\,\mathrm{m/s}, F=1\,\mathrm{N}, \theta=90^{\circ}.

Page 10 – Motion in a Uniform Electric Field

  • Particle (+q, mass m) enters region length x with initial horizontal speed v, field E\vec E vertical.
  • Force qEqE upward → acceleration a=qE/ma=qE/m.
  • Time inside field t=x/vt=x/v.
  • Vertical displacement y=12at2=qEx22mv2y = \tfrac{1}{2} a t^{2} = \frac{qE x^{2}}{2 m v^{2}}.
  • Trajectory is a parabola: x2=2mv2qEyx^{2} = \frac{2m v^{2}}{qE}\, y.

Page 11 – Motion in a Uniform Magnetic Field

  • Velocity components: v<em>=vcosθv<em>{\parallel}=v \cos\theta (along B\vec B), v</em>=vsinθv</em>{\perp}=v \sin\theta (perpendicular).
  • Perpendicular component causes circular motion (centripetal force from magnetic Lorentz force):
    mv<em>2r=qv</em>Br=mvsinθqB\frac{m v<em>{\perp}^{2}}{r} = q v</em>{\perp} B \Rightarrow r = \frac{m v \sin\theta}{q B}.
  • Angular (cyclotron) frequency and time period:
    ω=qBm,T=2πmqB\omega = \frac{qB}{m},\qquad T = \frac{2\pi m}{qB} (independent of speed and radius).
  • Parallel component → uniform motion → helix; pitch (advance per turn):
    p=vcosθ  T=2πmvcosθqBp = v \cos\theta \; T = \frac{2\pi m v \cos\theta}{qB}.

Page 12 – Lorentz Force in Combined E\vec E and B\vec B Fields

  • General force: F=q(E+v×B)\vec F = q(\vec E + \vec v \times \vec B).
  • Special situations
    1. E,B,v\vec E\, ,\vec B, \vec v collinear → magnetic part zero, particle accelerates linearly by E\vec E.
    2. EB\vec E\perp\vec B and v\vec v adjusted so that qE=qvBqE = qvB (opposite directions) → net force zero → particle passes undeflected. Velocity selector condition: v=E/Bv = E/B.

Page 13 – Force on a Current-Carrying Conductor

  • Consider conductor length ll carrying current II at angle θ\theta to field B\vec B.
  • Force: F=I(l×B)\vec F = I \,(\vec l \times \vec B); magnitude F=BIlsinθF = B I l \sin\theta.
  • Microscopic view: current I=neAvdI = n e A v_d (drift). Using NN electrons segment, force derives to same macroscopic form.
  • Direction by Fleming’s left-hand rule or right-hand palm rule.

Page 15 – Force Between Two Parallel Currents & Definition of Ampere

  • Currents I<em>1,I</em>2I<em>1, I</em>2 separated by distance dd.
  • Field from wire 1 at wire 2: B<em>1=μ</em>0I1/(2πd)B<em>1 = \mu</em>0 I_1 / (2\pi d).
  • Force per unit length on wire 2: Fl=I<em>2B</em>1=μ<em>0I</em>1I22πd\frac{F}{l}= I<em>2 B</em>1 = \frac{\mu<em>0 I</em>1 I_2}{2\pi d}.
    • Same magnitude (Newton’s 3rd), opposite direction.
  • Currents same direction → attractive; opposite → repulsive.
  • Ampere (definition): currents of 1A1\,\mathrm{A} in two infinite parallel conductors 1 m apart produce force 2×107  Nm12 \times 10^{-7}\;\mathrm{N\,m^{-1}}.

Page 17 – Torque on a Current Loop; Magnetic Dipole Moment

  • Rectangular loop (area A=lbA=lb) in uniform field B\vec B; angle between area vector A\vec A and B\vec B is θ\theta.
  • Forces on opposite sides produce couple; net torque:
    τ=BIAsinθ\tau = B I A \sin\theta.
  • For N turns: τ=BIANsinθ\tau = B I A N \sin\theta.
  • Define magnetic dipole moment:
    m=INA\vec m = I N \vec A.
    Then τ=m×B\vec \tau = \vec m \times \vec B, τmax=BIAN|\tau_{\max}| = B I A N (when loop plane perpendicular to field).

Page 18 – Moving-Coil Galvanometer (MCG)

  • Principle: current-carrying coil in radial magnetic field experiences torque τ=BINA\tau = B I N A (since sin90=1\sin 90^{\circ}=1).
  • Construction Elements
    1. Rectangular multi-turn coil, suspended by phosphor-bronze strip.
    2. Cylindrical pole pieces + soft-iron core → strong radial field (lines radial, plane of coil always parallel to field).
    3. Mirror or pointer attached; torsion spring provides restoring torque τr=kϕ\tau_r = k \phi.
  • Working equilibrium: BNAI=kϕI=(k/BNA)ϕ=GϕB N A I = k \phi \Rightarrow I = (k/ B N A) \phi = G \phi.
    • Galvanometer constant (figure of merit) G=k/(BNA)  A/divG = k / (B N A) \;\mathrm{A/div}.
  • Sensitivities
    1. Current sensitivity ϕ/I=BNA/k\phi/I = B N A / k (↑ by ↑N, B, A or ↓k).
    2. Voltage sensitivity ϕ/V=(ϕ/I)/Rg\phi/V = (\phi/I)/R_g (total series resistance).

Page 20 – Converting Galvanometer to Ammeter

  • Need low overall resistance; connect shunt SS parallel to galvanometer (resistance GG).
  • Let IgI_g be full-scale galvanometer current, desired full-scale II.
  • Same potential across parallel paths: (II<em>g)S=I</em>gG(I-I<em>g) S = I</em>g G
    S=I<em>gGII</em>gS = \frac{I<em>g G}{I - I</em>g} (≪ G).
  • Resulting ammeter resistance RA=GSG+SSR_A = \frac{G S}{G + S} \approx S (low).

Page 21 – Converting Galvanometer to Voltmeter

  • Need high resistance; connect series resistor RsR_s with galvanometer.
  • For full-scale voltage VV:
    V=I<em>g(G+R</em>s)R<em>s=VI</em>gGV = I<em>g (G + R</em>s) \Rightarrow R<em>s = \frac{V}{I</em>g} - G.
  • Voltmeter resistance R<em>V=G+R</em>sR<em>V = G + R</em>s (≫ circuit resistance to minimize loading).
  • Ideal voltmeter: R<em>VR<em>V \to \infty; ideal ammeter: R</em>A0R</em>A \to 0.

Page 21 (addendum) – Magnetic Dipole Moment of Bar Magnet

  • For magnet poles +p<em>m,p</em>m+p<em>m, -p</em>m separated by length 2l2l:
    M=pm(2l)M = p_m (2l) directed from south to north pole.
  • For current loop, same M=IANM = I A N (equivalence principle).