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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 B⃗\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

F⃗=qv⃗×B⃗\vec{F} = q\vec{v} \times \vec{B}

Magnitude: F=qvBsin⁡θF = qvB \sin\theta

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

Direction: Right-Hand Rule #1

  1. Point fingers along v⃗\vec{v} (velocity)
  2. Curl fingers toward B⃗\vec{B} (field)
  3. Thumb points along F⃗\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 v⃗\vec{v} and B⃗\vec{B}
  • No work done (F ⊥ v, so F·v = 0)

Circular Motion in B Field

If v⃗⊥B⃗\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!

F⃗=IL⃗×B⃗\vec{F} = I\vec{L} \times \vec{B}

Magnitude: F=BILsin⁡θF = 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 B⃗\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×10−7\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=NIABsin⁡θ\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=μBsin⁡θ\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=−1.6×10−19q = -1.6 \times 10^{-19} C
  • Mass: m=9.11×10−31m = 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=∣q∣vBsin⁡θ=(1.6×10−19)(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=1.6×10−13 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=mv∣q∣B=(9.11×10−31)(2.0×106)(1.6×10−19)(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=1.82×10−248.0×10−20=2.3×10−5 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 2easy

❓ 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=−1.6×10−19q = -1.6 \times 10^{-19} C
  • Mass: m=9.11×10−31m = 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=∣q∣vBsin⁡θ=(1.6×10−19)(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=1.6×10−13 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=mv∣q∣B=(9.11×10−31)(2.0×106)(1.6×10−19)(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=1.82×10−248.0×10−20=2.3×10−5 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

3Problem 3medium

❓ 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=BILsin⁡θ=(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

4Problem 4medium

❓ 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=BILsin⁡θ=(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

5Problem 5medium

❓ Question:

An electron (q = -1.6 × 10⁻¹⁹ C) moves at 2.0 × 10⁶ m/s perpendicular to a magnetic field of 0.50 T. (a) What is the magnitude of the magnetic force? (b) What is the direction of the force? (c) What is the radius of the circular path?

💡 Show Solution

Solution:

Given: q = -1.6 × 10⁻¹⁹ C, v = 2.0 × 10⁶ m/s, B = 0.50 T, m_e = 9.1 × 10⁻³¹ kg

(a) Magnetic force: F = |q|vB sin θ = |q|vB (since θ = 90°) F = (1.6 × 10⁻¹⁹)(2.0 × 10⁶)(0.50) F = 1.6 × 10⁻¹³ N

(b) Direction: Use right-hand rule: Point fingers in direction of v, curl toward B, thumb points in direction of F for positive charge. For electron (negative), force is opposite to right-hand rule direction.

(c) Radius of circular path: F = mv²/r (centripetal force) r = mv²/(|q|vB) = mv/(|q|B) r = (9.1 × 10⁻³¹)(2.0 × 10⁶)/[(1.6 × 10⁻¹⁹)(0.50)] r = (1.82 × 10⁻²⁴)/(8.0 × 10⁻²⁰) r = 2.3 × 10⁻⁵ m or 0.023 mm

6Problem 6hard

❓ 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=5.0×10−3 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=NIABsin⁡θ\tau = NIAB \sin\theta τ=(20)(3.0)(5.0×10−3)(0.40)sin⁡(60°)\tau = (20)(3.0)(5.0 \times 10^{-3})(0.40)\sin(60°) τ=(20)(3.0)(5.0×10−3)(0.40)(0.866)\tau = (20)(3.0)(5.0 \times 10^{-3})(0.40)(0.866) τ=0.052 N⋅m\tau = 0.052 \text{ N·m}

Answer: τ = 0.052 N·m = 52 mN·m

(Torque tends to align loop with field)

7Problem 7medium

❓ Question:

A wire carrying 5.0 A of current has length 0.30 m in a uniform 0.80 T magnetic field. The wire makes an angle of 60° with the field. (a) What is the magnetic force on the wire? (b) What angle gives maximum force?

💡 Show Solution

Solution:

Given: I = 5.0 A, L = 0.30 m, B = 0.80 T, θ = 60°

(a) Magnetic force: F = BIL sin θ F = (0.80)(5.0)(0.30) sin 60° F = (0.80)(5.0)(0.30)(0.866) F = 1.04 N or 1.0 N

(b) Maximum force angle: F is maximum when sin θ = 1 θ = 90° (wire perpendicular to field)

8Problem 8hard

❓ 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=5.0×10−3 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=NIABsin⁡θ\tau = NIAB \sin\theta τ=(20)(3.0)(5.0×10−3)(0.40)sin⁡(60°)\tau = (20)(3.0)(5.0 \times 10^{-3})(0.40)\sin(60°) τ=(20)(3.0)(5.0×10−3)(0.40)(0.866)\tau = (20)(3.0)(5.0 \times 10^{-3})(0.40)(0.866) τ=0.052 N⋅m\tau = 0.052 \text{ N·m}

Answer: τ = 0.052 N·m = 52 mN·m

(Torque tends to align loop with field)

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