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Magnetic Fields and Forces

Lorentz force, motion of charges in magnetic fields

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Magnetic Fields and Forces

Magnetic Force on Moving Charge

Lorentz force: 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

Properties:

  • Force perpendicular to both v⃗\vec{v} and B⃗\vec{B}
  • Magnetic force does no work (F⃗⊥v⃗\vec{F} \perp \vec{v})
  • Changes direction, not speed

Circular Motion in Magnetic Field

For v⃗⊥B⃗\vec{v} \perp \vec{B}:

qvB=mv2rqvB = \frac{mv^2}{r}

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

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

Cyclotron frequency: f=qB2πmf = \frac{qB}{2\pi m}

(Independent of vv and rr!)

Helical Motion

If v⃗\vec{v} has components both parallel and perpendicular to B⃗\vec{B}:

  • Perpendicular component causes circular motion
  • Parallel component causes uniform motion
  • Result: helix

Pitch of helix: p=v∥T=2πmv∥qBp = v_{\parallel}T = \frac{2\pi m v_{\parallel}}{qB}

Combined Electric and Magnetic Fields

Total force: F⃗=q(E⃗+v⃗×B⃗)\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})

Velocity selector:

Crossed E⃗\vec{E} and B⃗\vec{B} perpendicular to v⃗\vec{v}:

Particles with v=E/Bv = E/B pass straight through (F⃗=0\vec{F} = 0).

Magnetic Force on Current

Current II in wire of length L⃗\vec{L} in field B⃗\vec{B}:

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

For straight wire: F=ILBsin⁡θF = ILB\sin\theta

For curved wire: F⃗=∫I dl⃗×B⃗\vec{F} = \int I \, d\vec{l} \times \vec{B}

Torque on Current Loop

Rectangular loop with area AA, current II, in uniform field B⃗\vec{B}:

Magnetic dipole moment: μ⃗=IAn^\vec{\mu} = IA\hat{n}

(where n^\hat{n} is normal to loop, by right-hand rule)

Torque: τ⃗=μ⃗×B⃗\vec{\tau} = \vec{\mu} \times \vec{B}

τ=μBsin⁡θ=IABsin⁡θ\tau = \mu B\sin\theta = IAB\sin\theta

Potential energy: U=−μ⃗⋅B⃗=−μBcos⁡θU = -\vec{\mu} \cdot \vec{B} = -\mu B\cos\theta

(Minimum when μ⃗\vec{\mu} parallel to B⃗\vec{B})

Applications

Mass spectrometer: Separates ions by mass using circular motion in B⃗\vec{B}

Cyclotron: Accelerates particles using constant frequency RF field

Hall effect: Voltage across conductor perpendicular to current and B⃗\vec{B}

VH=IBnqtV_H = \frac{IB}{nqt}

where nn is charge carrier density, tt is thickness.

📚 Practice Problems

1Problem 1easy

❓ Question:

A proton (q = 1.6 × 10⁻¹⁹ C, m = 1.67 × 10⁻²⁷ kg) enters a uniform magnetic field B = 0.5 T with velocity v = 2.0 × 10⁶ m/s perpendicular to the field. Find: (a) the magnitude of the magnetic force, (b) the radius of the circular path, and (c) the period of the motion.

💡 Show Solution

Given:

  • q = 1.6 × 10⁻¹⁹ C
  • m = 1.67 × 10⁻²⁷ kg
  • B = 0.5 T
  • v = 2.0 × 10⁶ m/s
  • θ = 90° (perpendicular)

(a) Magnetic force:

F=qvBsin⁡θ=(1.6×10−19)(2.0×106)(0.5)(1)F = qvB\sin\theta = (1.6 \times 10^{-19})(2.0 \times 10^6)(0.5)(1)

F=1.6×10−13 NF = \boxed{1.6 \times 10^{-13} \text{ N}}

(b) Radius of path:

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

r=mvqB=(1.67×10−27)(2.0×106)(1.6×10−19)(0.5)r = \frac{mv}{qB} = \frac{(1.67 \times 10^{-27})(2.0 \times 10^6)}{(1.6 \times 10^{-19})(0.5)}

r=4.18×10−2 m=4.18 cmr = \boxed{4.18 \times 10^{-2} \text{ m} = 4.18 \text{ cm}}

(c) Period:

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

T=2π(1.67×10−27)(1.6×10−19)(0.5)T = \frac{2\pi (1.67 \times 10^{-27})}{(1.6 \times 10^{-19})(0.5)}

T=1.31×10−7 s=131 nsT = \boxed{1.31 \times 10^{-7} \text{ s} = 131 \text{ ns}}

Note: Period is independent of velocity!

2Problem 2medium

❓ Question:

A rectangular current loop (dimensions: 0.2 m × 0.3 m) carries current I = 5.0 A. The loop is placed in a uniform magnetic field B = 0.8 T with the field parallel to the plane of the loop, making angle θ = 30° with the normal to the loop. Find: (a) the magnetic dipole moment, (b) the torque on the loop, and (c) the orientation angle for maximum torque.

💡 Show Solution

Given:

  • Dimensions: 0.2 m × 0.3 m
  • I = 5.0 A
  • B = 0.8 T
  • θ = 30° (angle between normal and field)

(a) Magnetic dipole moment:

μ=IA=I(length×width)\mu = IA = I(\text{length} \times \text{width})

μ=(5.0)(0.2)(0.3)=0.3 A⋅m2\mu = (5.0)(0.2)(0.3) = \boxed{0.3 \text{ A·m}^2}

Direction: perpendicular to loop (right-hand rule)

(b) Torque:

τ=μBsin⁡θ=(0.3)(0.8)sin⁡(30°)\tau = \mu B \sin\theta = (0.3)(0.8)\sin(30°)

τ=(0.24)(0.5)=0.12 N⋅m\tau = (0.24)(0.5) = \boxed{0.12 \text{ N·m}}

Direction: tends to align μ with B

(c) Maximum torque:

Torque is maximum when sin⁡θ=1\sin\theta = 1, i.e., when:

θ=90°\theta = \boxed{90°}

This occurs when the normal to the loop is perpendicular to B, meaning the magnetic field is in the plane of the loop.

Maximum torque: τmax=μB=(0.3)(0.8)=0.24 N⋅m\tau_{max} = \mu B = (0.3)(0.8) = 0.24 \text{ N·m}

3Problem 3hard

❓ Question:

A particle with charge q = 2.0 μC and mass m = 1.0 × 10⁻⁸ kg enters a region with perpendicular electric field E = 500 V/m and magnetic field B = 0.1 T. The particle moves in a straight line. Find: (a) the velocity of the particle (velocity selector), (b) if E is then turned off, find the radius of the resulting circular path, and (c) explain how this setup could be used as a mass spectrometer.

💡 Show Solution

Given:

  • q = 2.0 μC = 2.0 × 10⁻⁶ C
  • m = 1.0 × 10⁻⁸ kg
  • E = 500 V/m
  • B = 0.1 T
  • Straight-line motion initially

(a) Velocity (velocity selector):

For straight-line motion, electric and magnetic forces balance: FE=FBF_E = F_B qE=qvBqE = qvB

v=EB=5000.1v = \frac{E}{B} = \frac{500}{0.1}

v=5000 m/sv = \boxed{5000 \text{ m/s}}

This velocity is independent of q and m!

(b) Radius with E off:

With only magnetic force: r=mvqB=(1.0×10−8)(5000)(2.0×10−6)(0.1)r = \frac{mv}{qB} = \frac{(1.0 \times 10^{-8})(5000)}{(2.0 \times 10^{-6})(0.1)}

r=5.0×10−52.0×10−7r = \frac{5.0 \times 10^{-5}}{2.0 \times 10^{-7}}

r=0.25 m=25 cmr = \boxed{0.25 \text{ m} = 25 \text{ cm}}

(c) Mass spectrometer principle:

Since r=mvqBr = \frac{mv}{qB} and v=E/Bv = E/B:

r=m(E/B)qB=mEqB2r = \frac{m(E/B)}{qB} = \frac{mE}{qB^2}

For particles with same q but different m: r1r2=m1m2\frac{r_1}{r_2} = \frac{m_1}{m_2}

How it works:

  1. Velocity selector ensures all particles have same v (regardless of m)
  2. Magnetic field separates particles by mass
  3. Heavier particles follow larger radius paths
  4. Detector measures radius → determines mass
  5. Can separate isotopes (same q, different m)

Example: ²³⁵U and ²³⁸U ions would separate into different paths.

Explain using:
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