Motion of Charged Particles in Magnetic Fields

Motion of Charged Particles in Magnetic Fields

Magnetic Force

• Magnets, magnetic materials, moving charges, and current-carrying conductors experience a force in a magnetic field.

• The magnetic force on a moving charged particle depends on the magnitude and direction of the velocity of the moving charge.

• The direction of the force on a current-carrying conductor or charged particle moving at an angle $0$ to a uniform magnetic field depends on the direction of the magnetic field and the direction of charge movement.

• Determine the direction of one of the forces (magnetic field or charge movement) given the direction of the other two.

• Solve problems using $F = IlB \sin \theta$ for a current-carrying conductor and $F = qvB \sin \theta$ for a moving charged particle.

• The magnitude of the magnetic force on a current-carrying conductor depends on the current, magnetic field strength, the length of the conductor inside the magnetic field, and the angle between the current and magnetic field.

Formula for Magnetic Force on a Current-Carrying Conductor

$F = IlB \sin \theta$

• Where:

  • $F$ = Magnetic force (N)

  • $I$ = Current (A)

  • $l$ = Length of the conductor in the magnetic field (m)

  • $B$ = Magnetic field strength (T)

  • $\theta$ = Angle between the current and magnetic field

Example Calculation

A 65 cm length of wire carries a current of 4A at an angle of 30° to a 2.2 mT magnetic field. Calculate the magnetic force on the wire:

$F = 4 \times 0.65 \times 2.2 \times 10^{-3} \times \sin 30$

$F = 2.9 \times 10^{-3} N$

Right-Hand Palm Rule

• The direction of the magnetic force on a current-carrying conductor is determined using the right-hand palm rule.

• When the right thumb points in the direction of conventional current and the fingers point in the direction of the magnetic field, the open palm pushes in the direction of the magnetic force.

Right-hand and Left-hand Rules

• Right-hand rule for electrically positively charged particles

  • The thumb points in the direction of the velocity ($v$) of the electrically positively charged particle.

  • The fingers point in the direction of the Magnetic field ($B$) (from North to South).

  • The palm faces the direction of the Force ($F$) on the electrically positively charged particle.

• Left-hand rule for electrically negatively charged particles

  • The thumb points in the direction of the velocity ($v$) of the electrically negatively charged particle.

  • The fingers point in the direction of the Magnetic field ($B$) (from North to South).

  • The palm faces the direction of the Force ($F$) on the electrically negatively charged particle.

Worked Example

A straight conductor carries a 2.0 A current perpendicular to a 0.2 T magnetic field. Calculate the magnitude and determine the direction of the force on the conductor:

$F = 2.0 \times 0.5 \times 0.2 \times \sin 90$

$F = 0.2 N$, vertically upwards

Worked Example 2

A straight conductor carries a 3.5 A current that is 60° to a 0.5 T magnetic field. Calculate the magnitude and determine the direction of the force on a 40 cm length of the conductor.

$F = 3.5 \times 0.4 \times 0.5 \times \sin 60$

$F = 0.6 N$, into the page

Magnetic Force on a Charged Particle

• A charged particle experiences a magnetic force when it moves in a magnetic field.

• The magnitude of the magnetic force on a moving charge depends on the velocity and charge of the particle, the magnetic field strength, and the angle between the velocity and magnetic field.

Formula for Magnetic Force on a Moving Charged Particle

$F = qvB \sin \theta$

• Where:

  • $F$ = Magnetic force (N)

  • $q$ = Charge on particle (C)

  • $v$ = Velocity of particle (m/s)

  • $B$ = Magnetic field strength (T)

  • $\theta$ = Angle between the field and velocity (°)

Direction of Magnetic Force

• The direction of the magnetic force on a moving charge is determined using the right-hand palm rule.

• For a positive charge, the direction of the magnetic force is the open push of the palm when the right thumb points in the direction of the velocity and the fingers point in the direction of the magnetic field.

• For a negative charge, the direction of magnetic force is obtained using the right-hand palm rule and reversing the outcome, or by using the left hand.

Worked Example

A proton moving at $1 \times 10^5$ m/s enters perpendicular to a 0.4T magnetic field. Calculate the magnitude and determine the direction of the force on the proton.

$F = 1.6 \times 10^{-19} \times 1 \times 10^5 \times 0.4 \times \sin 90$

$F = 6.4 \times 10^{-15} N$, vertically downwards

Worked Example 2

An electron moving at $2 \times 10^6$ m/s enters perpendicular to a 0.5T magnetic field. Calculate the magnitude and determine the direction of the force on the electron.

$F = 1.6 \times 10^{-19} \times 2 \times 10^6 \times 0.5 \times \sin 90$

$F = 1.6 \times 10^{-13} N$, vertically upwards

Charged Particles Circular Motion

• A charged particle moving at right angles to a uniform magnetic field experiences a force of constant magnitude at right angles to the velocity. The force changes the direction but not the speed of the charged particle; hence, the particle moves with uniform circular motion.

• Explain how the velocity dependence of the magnetic force on a charged particle causes the particle to move with uniform circular motion when it enters a uniform magnetic field at right angles.

• Derive $r = \frac{mv}{Bq}$ for the radius of the circular path of an ion of charge $q$ and mass $m$ that is moving with speed $v$ at right angles to a uniform magnetic field of magnitude $B$.

• Solve problems involving the use of $r = \frac{mv}{Bq}$.

• When moving perpendicular to a uniform magnetic field, a moving charge experiences a magnetic force at right angles to its velocity.

• The magnetic force causes the charged particle to move with uniform circular motion inside the field.

• A formula for the radius of the circular path is derived by asserting that the magnetic force causes the centripetal acceleration of a charge moving perpendicular to a uniform magnetic field.

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

(cancel $v$)

$\frac{mv}{r} = qB$

$r = \frac{mv}{qB}$

Radius of a Circular Path of a Moving Charge

$r = \frac{mv}{qB}$

• Where:

  • $r$ = Radius of the circular path (m)

  • $m$ = Mass of particle (kg)

  • $v$ = Velocity of particle (m/s)

  • $q$ = Charge of particle (C)

  • $B$ = Magnetic field strength (T)

Worked Example

An electron moving at $4 \times 10^7$ m/s enters perpendicular to 2 mT magnetic field. Calculate the radius of the circular path of the electron inside the field.

$r = \frac{mv}{qB}$

$r = \frac{9.11 \times 10^{-31} \times 4 \times 10^7}{1.6 \times 10^{-19} \times 2 \times 10^{-3}}$

$r = 0.11 m$

Circular Path of Charged Particles

• Cyclotrons are used to accelerate ions to high speed. Radioisotopes used in medicine and industry may be produced from collisions between high-speed ions and nuclei.

• The magnetic field within the dees of a cyclotron causes the charged particles to travel in a circular path so that they repeatedly pass through the electric field.

• Describe the nature and direction of the magnetic field needed to deflect ions into a circular path in the dees of a cyclotron.

• Derive the formula $T = \frac{2\pi m}{qB}$ for the period $T$ of the circular motion of an ion and hence show that the period is independent of the speed of the ion.

• Derive the formula $E<em>k = \frac{q^2 B^2 r^2}{2m}$ for the kinetic energy $E</em>k$ of the ions emerging at radius $r$ from a cyclotron.

• Explain why $E_k$ is independent of the potential difference across the dees and, for given ions, depends only on the magnetic field and the radius of the cyclotron.

• Solve problems involving the use of $T = \frac{2\pi m}{qB}$ and $E_k = \frac{q^2 B^2 r^2}{2m}$.

Particle Accelerators

• Recall that a cyclotron is a particle accelerator that accelerates ions to a high speed.

• They are used to produce radioisotopes for medical and industrial applications.

• Ions must be moving at high speed for a reaction to occur since both the ion and target nuclei are positively charged and must overcome the electrostatic force of repulsion.

Period of Circular Motion

• A cyclotron contains two electromagnets positioned above and below the plane of the dees that produce a permanent and uniform magnetic field.

• The magnetic field exerts a force that effectively steers the ions repeatedly in the direction of the electric field, which does work on the ions and increases their speed.

• The radius increases with the speed of the ion, but the period of circular motion is constant as the increase in speed is offset by the increase in the length of the circular path.

• The period is independent of the speed of the ion.

Derivation of the Period of Circular Motion

$v = \frac{qBr}{m}$

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi r}{\frac{qBr}{m}}$

(cancel $r$ and multiply by $m$)

$\frac{2\pi m}{T} = qB$

$T = \frac{2\pi m}{qB}$

$T = \frac{2\pi m}{qB}$

• Where:

  • $T$ = Period of circular motion (s)

  • $m$ = Mass of ion (kg)

  • $q$ = Charge of ion (C)

  • $B$ = Magnetic field strength (T)

Worked Example

A proton is accelerated by a cyclotron with a 0.8 T magnetic field. Calculate the period of the circular motion of the proton.

$T = \frac{2\pi m}{qB}$

$T = \frac{2\pi \times 1.673 \times 10^{-27}}{1.6 \times 10^{-19} \times 0.8}$

$T = 8.2 \times 10^{-8} s$

Kinetic Energy of Emerging Ions

• The ion's speed, kinetic energy, and radius increase each time they cross the gap between the dees.

• The ions emerge from the cyclotron when the radius of the circular path is equal to the internal radius of the cyclotron.

• The kinetic energy of the emerging ions is dependent only on the radius and magnetic field strength of the cyclotron.

• The particle’s mass and charge are independent of the potential difference across the dees.

Derivation of Kinetic Energy of Emerging Ions

$E_k = \frac{1}{2}mv^2$

$v = \frac{qBr}{m}$

$E_k = \frac{1}{2}m(\frac{qBr}{m})^2$

$E_k = \frac{q^2 B^2 r^2}{2m}$

$E_k = \frac{q^2 B^2 r^2}{2m}$

• Where:

  • $E_k$ = Kinetic energy of emerging ions (J)

  • $q$ = Charge of ion (C)

  • $B$ = Magnetic field strength (T)

  • $r$ = Radius at which ions emerge (m)

  • $m$ = Mass of ion (kg)

Worked Example

Calculate the kinetic energy of protons emerging from a cyclotron with a radius of 75 cm and a 0.9 T magnetic field within the dees.

$E_k = \frac{q^2 B^2 r^2}{2m}$

$E_k = \frac{(1.6 \times 10^{-19})^2 (0.9)^2 (0.75)^2}{2(1.673 \times 10^{-27})}$

$E_k = 3.5 \times 10^{-12} J$