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:
  • When an electron enters a magnetic field from the left, and if 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