Magnetic Fields and Forces

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

🧲 Magnetic Fields and Forces

Magnetic Field

Magnetic field $\vec{B}$ 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

Permanent magnets: Aligned atomic magnetic moments
Moving charges: Create B field
Current-carrying wires: Moving charges → B field
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

$\vec{F} = q\vec{v} \times \vec{B}$

Magnitude: $F = qvB \sin\theta$

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

Direction: Right-Hand Rule #1

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

For negative charge: Force is opposite!

Key Points:

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

Circular Motion in B Field

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

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

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

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

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

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

Applications: Mass Spectrometer

Separates ions by mass:

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

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

Force on Current-Carrying Wire

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

$\vec{F} = I\vec{L} \times \vec{B}$

Magnitude: $F = BIL \sin\theta$

where:

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

Direction: Right-Hand Rule #2

Point fingers along current direction
Curl toward $\vec{B}$
Thumb = force direction

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

Force Between Parallel Wires

Two parallel wires carrying currents:

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

where:

$\mu_0 = 4\pi \times 10^{-7}$ 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:

$\tau = NIAB \sin\theta$

where:

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

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

$\tau = \mu B \sin\theta$

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

Applications: Electric Motor

Current through loop in B field
Torque rotates loop
Commutator reverses current every half-turn
Continuous rotation!

Magnetic Field of Long Straight Wire

$B = \frac{\mu_0 I}{2\pi r}$

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): $B = \mu_0 nI = \mu_0 \frac{N}{L} I$

where n = N/L = turns per unit length

Outside: B ≈ 0

Solenoid is like bar magnet with N and S poles!

Problem-Solving Strategy

Identify: Charge or current in B field?
Find angle $\theta$ between v (or I) and B
Calculate magnitude: F = qvB sinθ or F = BIL sinθ
Find direction: Right-hand rule
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: $q = -1.6 \times 10^{-19}$ C
Mass: $m = 9.11 \times 10^{-31}$ kg
Velocity: $v = 2.0 \times 10^6$ m/s
B field: $B = 0.50$ T
Angle: $\theta = 90°$ (perpendicular)

Part (a): Magnetic force

$F = |q|vB \sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50)(1)$ $F = 1.6 \times 10^{-13} \text{ N}$

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

Part (b): Radius of circular path

$r = \frac{mv}{|q|B} = \frac{(9.11 \times 10^{-31})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.50)}$ $r = \frac{1.82 \times 10^{-24}}{8.0 \times 10^{-20}} = 2.3 \times 10^{-5} \text{ m} = 0.023 \text{ mm}$

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

Part (a): Magnetic force

$F = |q|vB \sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.50)(1)$ $F = 1.6 \times 10^{-13} \text{ N}$

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

Part (b): Radius of circular path

Very small radius!

Answer:

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

3Problem 3medium

❓ 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

4Problem 4medium

❓ Question:

💡 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

5Problem 5medium

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

Solution:

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

Direction: Use right-hand rule #2

Fingers along current
Curl toward B field
Thumb = force direction

Answer: F = 2.4 N

6Problem 6medium

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

Solution:

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

Direction: Use right-hand rule #2

Fingers along current
Curl toward B field
Thumb = force direction

Answer: F = 2.4 N

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: $N = 20$
Current: $I = 3.0$ A
B field: $B = 0.40$ T
Angle: Plane at 30° to field

Solution:

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

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

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

Answer: τ = 0.052 N·m = 52 mN·m

(Torque tends to align loop with field)

9Problem 9hard

❓ 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: $N = 20$
Current: $I = 3.0$ A
B field: $B = 0.40$ T
Angle: Plane at 30° to field

Solution:

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

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

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

Answer: τ = 0.052 N·m = 52 mN·m

(Torque tends to align loop with field)

10Problem 10medium

❓ Question:

💡 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)

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