Magnetic Fields and Forces

Magnetic fields, forces on moving charges, forces on current-carrying wires, torque on current loops

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🧲 Magnetic Fields and Forces

Magnetic Field

Magnetic field B\vec{B} is a vector field created by moving charges.

Unit: Tesla (T) = N/(A·m) = kg/(A·s²)

Also: Gauss (G) where 1 T = 10⁴ G

Earth's field: ~0.5 G = 5 × 10⁻⁵ T


Sources of Magnetic Fields

  1. Permanent magnets: Aligned atomic magnetic moments
  2. Moving charges: Create B field
  3. Current-carrying wires: Moving charges → B field
  4. Electromagnets: Coils of wire with current

Magnetic Poles:

  • North and South (like charges, but no magnetic monopoles!)
  • Like poles repel, opposite poles attract
  • Field lines: N → S outside magnet

Force on Moving Charge

F=qv×B\vec{F} = q\vec{v} \times \vec{B}

Magnitude: F=qvBsinθF = qvB \sin\theta

where θ\theta is angle between v\vec{v} and B\vec{B}.

Direction: Right-Hand Rule #1

  1. Point fingers along v\vec{v} (velocity)
  2. Curl fingers toward B\vec{B} (field)
  3. Thumb points along F\vec{F} (force) for positive charge

For negative charge: Force is opposite!

Key Points:

  • θ=90°\theta = 90°: Maximum force (F=qvBF = qvB)
  • θ=0°\theta = 0° or 180°180°: No force (parallel or anti-parallel)
  • Force is perpendicular to both v\vec{v} and B\vec{B}
  • No work done (F ⊥ v, so F·v = 0)

Circular Motion in B Field

If vB\vec{v} \perp \vec{B}, charge moves in circle!

Magnetic force provides centripetal force: qvB=mv2rqvB = \frac{mv^2}{r}

Radius of orbit: r=mvqBr = \frac{mv}{qB}

Period: T=2πrv=2πmqBT = \frac{2\pi r}{v} = \frac{2\pi m}{qB}

Frequency (cyclotron frequency): f=1T=qB2πmf = \frac{1}{T} = \frac{qB}{2\pi m}

💡 Note: Period and frequency are independent of v and r!


Applications: Mass Spectrometer

Separates ions by mass:

  1. Ions accelerated through V
  2. Enter perpendicular B field
  3. Radius depends on m/q ratio
  4. Heavier ions have larger radius

r=1B2mVqr = \frac{1}{B}\sqrt{\frac{2mV}{q}}


Force on Current-Carrying Wire

Current = moving charges, so wire in B field experiences force!

F=IL×B\vec{F} = I\vec{L} \times \vec{B}

Magnitude: F=BILsinθF = BIL \sin\theta

where:

  • I = current (A)
  • L = length of wire in field (m)
  • θ\theta = angle between wire and B field

Direction: Right-Hand Rule #2

  1. Point fingers along current direction
  2. Curl toward B\vec{B}
  3. Thumb = force direction

Maximum force: Wire perpendicular to B (F=BILF = BIL) No force: Wire parallel to B


Force Between Parallel Wires

Two parallel wires carrying currents:

FL=μ0I1I22πd\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}

where:

  • μ0=4π×107\mu_0 = 4\pi \times 10^{-7} T·m/A (permeability of free space)
  • d = distance between wires

Same direction currents: Attract Opposite direction currents: Repel

💡 This defines the Ampere! 1 A = current that produces 2×10⁻⁷ N/m force between parallel wires 1 m apart.


Torque on Current Loop

Rectangular loop (sides a and b) in B field:

τ=NIABsinθ\tau = NIAB \sin\theta

where:

  • N = number of turns
  • I = current
  • A = area of loop
  • θ\theta = angle between normal to loop and B field

Magnetic dipole moment: μ=NIA\mu = NIA (A·m²)

τ=μBsinθ\tau = \mu B \sin\theta

Maximum torque: Loop perpendicular to B (θ=90°\theta = 90°) Equilibrium: Loop parallel to B (θ=0°\theta = 0°)


Applications: Electric Motor

  1. Current through loop in B field
  2. Torque rotates loop
  3. Commutator reverses current every half-turn
  4. Continuous rotation!

Magnetic Field of Long Straight Wire

B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}

Direction: Right-Hand Rule #3 (Grip Rule)

  • Thumb along current
  • Fingers curl around wire → B field direction

Field forms circles around wire.


Magnetic Field of Solenoid

Long coil of wire (N turns, length L):

Inside (uniform field): B=μ0nI=μ0NLIB = \mu_0 nI = \mu_0 \frac{N}{L} I

where n = N/L = turns per unit length

Outside: B ≈ 0

Solenoid is like bar magnet with N and S poles!


Problem-Solving Strategy

  1. Identify: Charge or current in B field?
  2. Find angle θ\theta between v (or I) and B
  3. Calculate magnitude: F = qvB sinθ or F = BIL sinθ
  4. Find direction: Right-hand rule
  5. For circular motion: Use r = mv/qB

Common Mistakes

❌ Forgetting sin θ term ❌ Wrong right-hand rule (for negative charges) ❌ Confusing E and B field formulas ❌ Saying magnetic force does work (it doesn't! F ⊥ v) ❌ Using v when parallel to B (F = 0) ❌ Wrong units (convert G to T, cm to m)

📚 Practice Problems

1Problem 1easy

Question:

An electron (q = -1.6×10⁻¹⁹ C, m = 9.11×10⁻³¹ kg) moves at 2.0×10⁶ m/s perpendicular to a 0.50 T magnetic field. Find (a) the magnetic force, (b) the radius of its circular path.

💡 Show Solution

Given:

  • Charge: q=1.6×1019q = -1.6 \times 10^{-19} C
  • Mass: m=9.11×1031m = 9.11 \times 10^{-31} kg
  • Velocity: v=2.0×106v = 2.0 \times 10^6 m/s
  • B field: B=0.50B = 0.50 T
  • Angle: θ=90°\theta = 90° (perpendicular)

Part (a): Magnetic force

F=qvBsinθ=(1.6×1019)(2.0×106)(0.50)(1)F = |q|vB \sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50)(1) F=1.6×1013 NF = 1.6 \times 10^{-13} \text{ N}

(Direction: Use right-hand rule, then reverse for negative charge)

Part (b): Radius of circular path

r=mvqB=(9.11×1031)(2.0×106)(1.6×1019)(0.50)r = \frac{mv}{|q|B} = \frac{(9.11 \times 10^{-31})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.50)} r=1.82×10248.0×1020=2.3×105 m=0.023 mmr = \frac{1.82 \times 10^{-24}}{8.0 \times 10^{-20}} = 2.3 \times 10^{-5} \text{ m} = 0.023 \text{ mm}

Very small radius!

Answer:

  • (a) F = 1.6 × 10⁻¹³ N
  • (b) r = 0.023 mm

2Problem 2medium

Question:

A straight wire carries 10 A current. A 0.30 m section lies perpendicular to a 0.80 T magnetic field. What is the magnetic force on this section?

💡 Show Solution

Given:

  • Current: I=10I = 10 A
  • Length: L=0.30L = 0.30 m
  • B field: B=0.80B = 0.80 T
  • Angle: θ=90°\theta = 90° (perpendicular)

Solution:

Force on current-carrying wire: F=BILsinθ=(0.80)(10)(0.30)(1)F = BIL \sin\theta = (0.80)(10)(0.30)(1) F=2.4 NF = 2.4 \text{ N}

Direction: Use right-hand rule #2

  • Fingers along current
  • Curl toward B field
  • Thumb = force direction

Answer: F = 2.4 N

3Problem 3hard

Question:

A rectangular loop (10 cm × 5 cm) with 20 turns carries 3.0 A current. It is placed in a 0.40 T field with its plane at 30° to the field. Find the torque on the loop.

💡 Show Solution

Given:

  • Dimensions: 10 cm × 5 cm
  • Number of turns: N=20N = 20
  • Current: I=3.0I = 3.0 A
  • B field: B=0.40B = 0.40 T
  • Angle: Plane at 30° to field

Solution:

Step 1: Find area. A=(0.10)(0.05)=0.0050 m2=5.0×103 m2A = (0.10)(0.05) = 0.0050 \text{ m}^2 = 5.0 \times 10^{-3} \text{ m}^2

Step 2: Find angle θ. If plane is at 30° to field, then normal to plane is at 60° to field! θ=90°30°=60°\theta = 90° - 30° = 60°

Step 3: Calculate torque. τ=NIABsinθ\tau = NIAB \sin\theta τ=(20)(3.0)(5.0×103)(0.40)sin(60°)\tau = (20)(3.0)(5.0 \times 10^{-3})(0.40)\sin(60°) τ=(20)(3.0)(5.0×103)(0.40)(0.866)\tau = (20)(3.0)(5.0 \times 10^{-3})(0.40)(0.866) τ=0.052 N\cdotpm\tau = 0.052 \text{ N·m}

Answer: τ = 0.052 N·m = 52 mN·m

(Torque tends to align loop with field)