5.4.2 Magnetic Force on a Charge

Calculating Magnetic Force on a Moving Charge

• The magnetic force on an isolating moving charge, such an electron, is given by the equation:

F = BQv sinθ

• Where:
  • F = force on the charge (N)
  • B = magnetic flux density (T)
  • Q = charge of the particle (C)
  • v = speed of the charge (m s^-1)
  • θ = angle between charge’s velocity and magnetic field (degrees)

The force on an isolated moving charge is perpendicular to its motion and the magnetic field B

• Equivalent to the force on a wire, if the magnetic field B is perpendicular to the direction of the charge’s velocity, the equation simplifies to:

F = BQv

• According to Fleming’s left hand rule:
  • If the electron velocity is to the left (and so the conventional current flows to the right), and the magnetic field is directed into the page, then the force on it will be directed upwards

• The equation shows:
  • If the direction of the electron changes, the magnitude of the force will change too

• The force due to the magnetic field is always perpendicular to the velocity of the electron
  • Note: this is equivalent to circular motion

• Fleming’s left-hand rule can be used again to find the direction of the force, magnetic field and velocity
  • The key difference is that the second finger representing current I (direction of positive charge) is now the direction of velocity v of the positive charge

The electron experiences a force upwards when it travels through the magnetic field between the two poles

Step 1: Write out the known quantities

Speed of the electron, v = 5.3 × 10⁷ m s^-1

Charge of an electron, Q = 1.60 × 10^-19 C

Magnetic flux density, B = 0.2 T

Angle between electron and magnetic field, θ = 30^°

Step 2: Write down the equation for the magnetic force on an isolated particle

F = BQv sinθ

Step 3: Substitute in values, and calculate the force on the electron at 30^°

F = (0.2) × (1.60 × 10^-19) × (5.3 × 10⁷) × sin(30) = 8.5 × 10^-13 N

Step 4: Calculate the electron force when travelling perpendicular to the field

F = BQv = (0.2) × (1.60 × 10^-19) × (5.3 × 10⁷) = 1.696 × 10^-12 N

Step 5: Calculate the ratio of the perpendicular force to the force at 30^°

Therefore, the force on the electron is twice as strong when it is moving perpendicular to the field than when it is moving at 30^° to the field

Motion of a Charged Particle in a Uniform Magnetic Field

• A charged particle in uniform magnetic field which is perpendicular to its direction of motion travels in a circular path
• This is because the magnetic force F_B will always be perpendicular to its velocity v
  • F_B will always be directed towards the centre of the path

A charged particle moves travels in a circular path in a magnetic field

• The magnetic force F_B provides the centripetal force on the particle
• Recall the equation for centripetal force:

• Where:
  • m = mass of the particle (kg)
  • v = linear velocity of the particle (m s^-1)
  • r = radius of the orbit (m)

• Equating this to the force on a moving charged particle gives the equation:

• Rearranging for the radius r obtains the equation for the radius of the orbit of a charged particle in a perpendicular magnetic field:

• This equation shows that:
  • Faster moving particles with speed v move in larger circles (larger r): r ∝ v
  • Particles with greater mass m move in larger circles: r ∝ m
  • Particles with greater charge q move in smaller circles: r ∝ 1 / q
  • Particles moving in a strong magnetic field B move in smaller circles: r ∝ 1 / B

Step 1: Write down the known quantities

Magnetic flux density, B = 6.2 mT

Electron speed, v = 3.0 × 10⁶ m s^-1

Step 2: Write down the equation for the radius of a charged particle in a perpendicular magnetic field

Step 3: Substitute in values