Magnetic Fields & Forces – Comprehensive Bullet‐Point Notes Magnetism Fundamentals Magnetism originates from moving electric charges (currents) and the intrinsic motion (spin) of sub-atomic particles.Unlike electrostatics, magnetostatic forces act only on moving charges. Two-stage interaction analogy:Stage 1 (Electric): A stationary charge creates an electric field E ⃗ \vec E E ; another charge feels F ⃗ E = q E ⃗ \vec F_E=q\vec E F E = q E . Stage 1 (Magnetic): A moving charge/current creates a magnetic field B ⃗ \vec B B . Stage 2: A second moving charge/current inside that B ⃗ \vec B B experiences a magnetic force. Permanent magnets contain micro-currents (electron circulation/spin) that mimic macroscopic currents. Magnetic poles come in pairs (no isolated magnetic ‘charges’ found ⇒ ∇ ! ⋅ B ⃗ = 0 \nabla!\cdot\vec B=0 ∇ ! ⋅ B = 0 ). Earth behaves like a huge bar magnet; the geographic North Pole is near a magnetic south pole.Field reversals occur irregularly (every 10 4 – 10 7 10^4–10^7 1 0 4 –1 0 7 yrs). Magnetic Force on a Single Charge Vector form: F ⃗ = q v ⃗ × B ⃗ \boxed{\vec F=q\,\vec v\times\vec B} F = q v × B .∣ F ⃗ ∣ = ∣ q ∣ v B sin θ |\vec F|=|q|vB\sin\theta ∣ F ∣ = ∣ q ∣ v B sin θ ((\theta) = angle between v ⃗ \vec v v and B ⃗ \vec B B ).SI unit of B ⃗ \vec B B : tesla (T) where 1 T = 1 N / ( A⋅m ) 1\,\text{T}=1\,\text{N}/(\text{A·m}) 1 T = 1 N / ( A⋅m ) . Direction via right-hand rule (RHR):Positive charge: fingers v ⃗ → \vec v\rightarrow v → B ⃗ \vec B B , thumb gives F ⃗ \vec F F . Negative charge: force is opposite thumb. Special orientations:θ = 0 ° \theta=0° θ = 0° or 180 ° 180° 180° : F ⃗ = 0 \vec F=0 F = 0 (parallel/anti-parallel motion).θ = 90 ° \theta=90° θ = 90° : ∣ F ⃗ ∣ = ∣ q ∣ v B |\vec F|=|q|vB ∣ F ∣ = ∣ q ∣ v B (maximum). Example 7.1 – Proton in uniform B ⃗ \vec B B
Data: q = + 1.6 × 10 − 19 C , v = 3.0 × 10 5 m/s , B = 2.0 T , θ = 30 ° q=+1.6\times10^{-19}\,\text{C},\ v=3.0\times10^{5}\,\text{m/s},\ B=2.0\,\text{T},\ \theta=30° q = + 1.6 × 1 0 − 19 C , v = 3.0 × 1 0 5 m/s , B = 2.0 T , θ = 30° . ∣ F ⃗ ∣ = ( 1.6 × 10 − 19 ) ( 3.0 × 10 5 ) ( 2.0 ) sin 30 ° = 4.8 × 10 − 14 N |\vec F|=(1.6\times10^{-19})(3.0\times10^{5})(2.0)\sin30°=4.8\times10^{-14}\,\text{N} ∣ F ∣ = ( 1.6 × 1 0 − 19 ) ( 3.0 × 1 0 5 ) ( 2.0 ) sin 30° = 4.8 × 1 0 − 14 N .RHR → force points in − y ^ -\hat y − y ^ ; sign flips for electrons. Magnetic Field Lines & Flux Field line properties:Tangent to B ⃗ \vec B B at every point. Density ∝ |B ⃗ \vec B B |. Emerge from N-poles, enter S-poles, and form closed loops (Gauss’s law for magnetism ∮ closed B ⃗ ⋅ d A ⃗ = 0 \displaystyle \oint_{\text{closed}}\vec B\cdot d\vec A=0 ∮ closed B ⋅ d A = 0 ). Magnetic flux through surface A A A : Φ < e m > B = ∫ ! B ⃗ ⋅ d A ⃗ = B < / e m > ⊥ A = B A cos θ \boxed{\Phi<em>B=\int!\vec B\cdot d\vec A=B</em>{\perp}A = BA\cos\theta} Φ < e m > B = ∫ ! B ⋅ d A = B < / e m > ⊥ A = B A cos θ .Unit: weber (Wb) where 1 Wb = 1 T⋅m 2 1\,\text{Wb}=1\,\text{T·m}^2 1 Wb = 1 T⋅m 2 . Example 7.2 – Flat plate
A = 3.0 cm 2 = 3.0 × 10 − 4 m 2 , Φ B = + 0.90 mWb A=3.0\,\text{cm}^2=3.0\times10^{-4}\,\text{m}^2,\ \Phi_B=+0.90\,\text{mWb} A = 3.0 cm 2 = 3.0 × 1 0 − 4 m 2 , Φ B = + 0.90 mWb .B = Φ B A cos 60 ° = 6.0 T B=\dfrac{\Phi_B}{A\cos60°}=6.0\,\text{T} B = A cos 60° Φ B = 6.0 T ; area vector makes 60 ° 60° 60° with B ⃗ \vec B B (not 120 ° 120° 120° because flux given +ve).Motion of a Charged Particle in Uniform B ⃗ \vec B B F ⃗ \vec F F ⟂ v ⃗ \vec v v → speed is constant; work done = 0.Pure perpendicular entry (cyclotron motion):Radius R = m v ∣ q ∣ B \displaystyle R=\frac{mv}{|q|B} R = ∣ q ∣ B m v . Angular speed ω = ∣ q ∣ B m \omega=\dfrac{|q|B}{m} ω = m ∣ q ∣ B ; cyclotron frequency f = ω / 2 π f=\omega/2\pi f = ω /2 π (mass-spectrometry & cyclotrons). Oblique entry: helical path (parallel component unchanged). Example 7.3 – Magnetron (microwave oven)
Required B B B so electrons orbit at f = 2450 MHz f=2450\,\text{MHz} f = 2450 MHz . ω = 2 π f = 1.54 × 10 10 s − 1 \omega=2\pi f=1.54\times10^{10}\,\text{s}^{-1} ω = 2 π f = 1.54 × 1 0 10 s − 1 .For electron m < e m > e = 9.11 × 10 − 31 kg , q = 1.60 × 10 − 19 C m<em>e=9.11\times10^{-31}\,\text{kg},\ q=1.60\times10^{-19}\,\text{C} m < e m > e = 9.11 × 1 0 − 31 kg , q = 1.60 × 1 0 − 19 C :
B = m < / e m > e ω ∣ q ∣ = 0.0877 T B=\dfrac{m</em>e\omega}{|q|}=0.0877\,\text{T} B = ∣ q ∣ m < / e m > e ω = 0.0877 T (easy with permanent magnet). Magnetic Force on Current-Carrying Conductors For straight segment length ℓ \ell ℓ carrying current I I I :F ⃗ = I ℓ ⃗ × B ⃗ \boxed{\vec F=I\,\vec \ell\times\vec B} F = I ℓ × B ((\vec \ell) points with current). Magnitude F = I ℓ B sin θ F=I\ell B\sin\theta F = I ℓ B sin θ ; RHR as before. Example 7.5 – Copper rod between electromagnet poles
I = 50.0 A ( west → east ) , B = 1.20 T I=50.0\,\text{A}\ (\text{west}→\text{east}),\ B=1.20\,\text{T} I = 50.0 A ( west → east ) , B = 1.20 T toward NE (45°).(a) F = I ℓ B sin 45 ° = 42.4 N F=I\ell B\sin45°=42.4\,\text{N} F = I ℓ B sin 45° = 42.4 N upward. (b) Max when rod rotated so θ = 90 ° \theta=90° θ = 90° ⇒ F max = 60.0 N F_{\max}=60.0\,\text{N} F m a x = 60.0 N (enables magnetic levitation if weight ≤60 N). Magnetic Torque on Current Loops & Coils Single planar loop:Magnetic moment (dipole): μ ⃗ = I A ⃗ \boxed{\vec \mu = I\,\vec A} μ = I A ((\vec A) ⟂ plane by RHR). Torque: τ ⃗ = μ ⃗ × B ⃗ \boxed{\vec \tau = \vec \mu \times \vec B} τ = μ × B ; magnitude τ = μ B sin θ = I A B sin θ \tau = \mu B\sin\theta = IAB\sin\theta τ = μ B sin θ = I A B sin θ . Potential energy: U = − μ ⃗ ⋅ B ⃗ = − μ B cos θ U=-\vec \mu\cdot\vec B=-\mu B\cos\theta U = − μ ⋅ B = − μ B cos θ . Coil with N N N tightly packed turns: μ = N I A , τ = N I A B sin θ , U = − N I A B cos θ \mu=NIA,\ \tau=N IAB\sin\theta,\ U=-NIA B\cos\theta μ = N I A , τ = N I A B sin θ , U = − N I A B cos θ . Example 7.6–7 (30-turn coil)
r = 0.0500 m , I = 5.00 A , B = 1.20 T , N = 30 r=0.0500\,\text{m},\ I=5.00\,\text{A},\ B=1.20\,\text{T},\ N=30 r = 0.0500 m , I = 5.00 A , B = 1.20 T , N = 30 (coil horizontal, θ = 90 ° \theta=90° θ = 90° initially).μ = N I A = 1.18 A⋅m 2 \mu=NIA=1.18\,\text{A·m}^2 μ = N I A = 1.18 A⋅m 2 .τ = μ B = 1.41 N⋅m \tau=\mu B=1.41\,\text{N·m} τ = μ B = 1.41 N⋅m .If coil rotates to align with B ⃗ ( θ = 0 ° ) \vec B\ (\theta=0°) B ( θ = 0° ) : Δ U = − 1.41 J \Delta U=-1.41\,\text{J} Δ U = − 1.41 J (energy released). Electric Motors (DC) Rotor = current loop; stator field exerts torque τ ⃗ = μ ⃗ × B ⃗ \vec \tau=\vec \mu\times\vec B τ = μ × B → mechanical rotation. Commutator reverses current every half-turn to maintain unidirectional torque. Back-emf ε \varepsilon ε induced by rotating rotor opposes supply (Lenz’s law). For series motor: V a b = ε + I r V_{ab}=\varepsilon + I r V ab = ε + I r . Power terms:Input P i n = V I P_{in}=VI P in = V I . Resistive loss P R = I 2 r P_R=I^2 r P R = I 2 r . Mechanical output P < e m > m e c h = P < / e m > i n − P R = ε I P<em>{mech}=P</em>{in}-P_R=\varepsilon I P < e m > m ec h = P < / e m > in − P R = ε I . Efficiency η = P < e m > m e c h / P < / e m > i n = ε / V \eta=P<em>{mech}/P</em>{in}=\varepsilon/V η = P < e m > m ec h / P < / e m > in = ε / V . Example 7.8 – 120-V motor, r = 2.00 Ω , I = 4.00 A r=2.00\,\Omega,\ I=4.00\,\text{A} r = 2.00 Ω , I = 4.00 A at full load
(a) ε = V − I r = 112 V \varepsilon=V-Ir=112\,\text{V} ε = V − I r = 112 V . (b) P i n = 480 W P_{in}=480\,\text{W} P in = 480 W . (c) P R = I 2 r = 32 W P_R=I^2 r=32\,\text{W} P R = I 2 r = 32 W . (d) P m e c h = 448 W P_{mech}=448\,\text{W} P m ec h = 448 W . (e) η = 93 % \eta=93\% η = 93% . (f) If rotor jams ⇒ ε → 0 \varepsilon→0 ε → 0 , current I = V / r = 60 A I=V/r=60\,\text{A} I = V / r = 60 A , losses P R = 7200 W P_R=7200\,\text{W} P R = 7200 W ⇒ catastrophic heating (fuses/breakers trip). The Hall Effect Charge carriers in conductor subject to B ⃗ \vec B B (perpendicular to current) experience magnetic deflection → transverse electric field E < e m > z E<em>z E < e m > z builds until q E < / e m > z = q v d B qE</em>z = qv_d B q E < / e m > z = q v d B . Hall voltage V < e m > H = E < / e m > z d V<em>H = E</em>z d V < e m > H = E < / e m > z d (thickness d d d of slab). Carrier concentration:n = J < e m > x B < / e m > y q E < e m > z \boxed{n = \dfrac{J<em>x B</em>y}{q E<em>z}} n = q E < e m > z J < e m > x B < / e m > y where J < / e m > x = I / A J</em>x=I/A J < / e m > x = I / A . Sign of V H V_H V H distinguishes electron vs hole conduction. Example 7.9 – Copper strip
Dimensions: thickness d = 2.0 mm , w = 1.50 cm d=2.0\,\text{mm},\ w=1.50\,\text{cm} d = 2.0 mm , w = 1.50 cm ; B = 0.40 T , I = 75 A , V H = 0.81 μ V B=0.40\,\text{T},\ I=75\,\text{A},\ V_H=0.81\,\mu\text{V} B = 0.40 T , I = 75 A , V H = 0.81 μ V . J < e m > x = 2.5 × 10 6 A/m 2 , E < / e m > z = 5.4 × 10 − 5 V/m J<em>x=2.5\times10^{6}\,\text{A/m}^2,\ E</em>z=5.4\times10^{-5}\,\text{V/m} J < e m > x = 2.5 × 1 0 6 A/m 2 , E < / e m > z = 5.4 × 1 0 − 5 V/m .n ≈ 1.16 × 10 29 m − 3 n\approx1.16\times10^{29}\,\text{m}^{-3} n ≈ 1.16 × 1 0 29 m − 3 (ideal free-electron model gives 8.5 × 10 28 m − 3 8.5\times10^{28}\,\text{m}^{-3} 8.5 × 1 0 28 m − 3 ).Magnetic Field of a Moving Point Charge Biot–Savart analogue for a single charge (steady velocity):
B ⃗ = μ 0 4 π q v ⃗ × r ^ r 2 \boxed{\vec B = \dfrac{\mu_0}{4\pi}\,\dfrac{q\,\vec v \times \hat r}{r^2}} B = 4 π μ 0 r 2 q v × r ^ .r ^ \hat r r ^ = unit vector from charge to field point, r r r = separation.Field circles around direction of motion (RHR #2: thumb = v ⃗ \vec v v , fingers give B ⃗ \vec B B ). Superposition applies for multiple charges/currents. Example 7.10 – Two protons moving oppositely along x x x
Electric repulsion: F < e m > E = 1 4 π ε < / e m > 0 q 2 r 2 F<em>E=\dfrac{1}{4\pi\varepsilon</em>0}\dfrac{q^2}{r^2} F < e m > E = 4 π ε < / e m > 0 1 r 2 q 2 upward on top proton. Magnetic interaction: lower proton’s B ⃗ \vec B B points + z +z + z at upper proton; force magnitudeF < e m > B = q v B = q v ( μ < / e m > 0 4 π q v r 2 ) = μ 0 q 2 v 2 4 π r 2 F<em>B=q v B=qv\left(\dfrac{\mu</em>0}{4\pi}\dfrac{qv}{r^2}\right)=\dfrac{\mu_0 q^2 v^2}{4\pi r^2} F < e m > B = q v B = q v ( 4 π μ < / e m > 0 r 2 q v ) = 4 π r 2 μ 0 q 2 v 2 downward (attractive) because currents oppose. Ratio F < e m > B F < / e m > E = μ < e m > 0 ε < / e m > 0 v 2 1 = ( v c ) 2 \displaystyle \frac{F<em>B}{F</em>E}=\frac{\mu<em>0\varepsilon</em>0 v^2}{1}=\left(\frac{v}{c}\right)^2 F < / e m > E F < e m > B = 1 μ < e m > 0 ε < / e m > 0 v 2 = ( c v ) 2 (tiny unless v ≈ c v\approx c v ≈ c ).Demonstrates why magnetic effects are relativistic corrections to Coulomb force. Practical & Conceptual Connections Magnetic levitation, MRI, mass spectrometers, cyclotrons, microwave magnetrons, Hall-effect sensors. Energy perspective: torque tendencies, potential minima (stable) vs maxima (unstable) for dipoles. Safety note: Motors draw huge stall currents; protective devices essential. Symmetry: Maxwell equation ∇ ! ⋅ B ⃗ = 0 \nabla!\cdot\vec B=0 ∇ ! ⋅ B = 0 implies closed field lines; no magnetic monopoles observed. Relativity link: Magnetic force can be viewed as electrostatic force in a different inertial frame (length contraction explains F B F_B F B scaling). Quick Reference – Key Equations Force on charge: F ⃗ = q ( v ⃗ × B ⃗ ) \vec F=q(\vec v\times\vec B) F = q ( v × B ) . Force on wire: F ⃗ = I ℓ ⃗ × B ⃗ \vec F=I\vec \ell\times\vec B F = I ℓ × B . Radius of circular motion: R = m v ∣ q ∣ B R=\dfrac{mv}{|q|B} R = ∣ q ∣ B m v . Magnetic moment: μ ⃗ = N I A ⃗ \vec \mu = N I \vec A μ = N I A . Torque: τ ⃗ = μ ⃗ × B ⃗ \vec \tau = \vec \mu \times \vec B τ = μ × B . Potential energy: U = − μ ⃗ ⋅ B ⃗ U=-\vec \mu\cdot\vec B U = − μ ⋅ B . Hall carrier density: n = J B q E H n=\dfrac{J B}{q E_H} n = q E H J B . Moving charge field: B ⃗ = μ 0 4 π q v ⃗ × r ^ r 2 \vec B = \dfrac{\mu_0}{4\pi} \dfrac{q\,\vec v \times \hat r}{r^2} B = 4 π μ 0 r 2 q v × r ^ .