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Motion of a Charged Particle in a Magnetic Field (CIE A Level Physics)

Revision Note

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Motion of a charged particle in a uniform magnetic field

  • When a charged particle enters a uniform magnetic field, it travels in a circular path
  • This is because the direction of the magnetic force F will always be
    • perpendicular to the particle's velocity v
    • directed towards the centre of the path, resulting in circular motion

Circular motion of a charge in a magnetic field

Circular motion of charged particle, downloadable AS & A Level Physics revision notes

In a magnetic field, a charged particle travels in a circular path as the force, velocity and field are all perpendicular

  • The magnetic force F provides the centripetal force on the particle
  • The equation for centripetal force is:

F space equals space fraction numerator m v squared over denominator r end fraction

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

fraction numerator m v squared over denominator r end fraction space equals space B Q v

  • Rearranging for the radius r gives an expression for the radius of the path of a charged particle in a perpendicular magnetic field:

r space equals space fraction numerator m v over denominator B Q end fraction

  • Where:
    • r = radius of the path (m)
    • m = mass of the particle (kg)
    • v = linear velocity of the particle (m s−1)
    • B = magnetic field strength (T)
    • Q = charge of the particle (C)
  • This equation shows that:
    • Faster moving particles with speed v move in larger circles (larger r):  r space proportional to space v
    • Particles with greater mass m move in larger circles:  r space proportional to space m
    • Particles with greater charge q move in smaller circles:  r space proportional to space 1 over q
    • Particles moving in a strong magnetic field B move in smaller circles:  r space proportional to space 1 over B
  • The centripetal acceleration is in the same direction as the magnetic (centripetal) force
  • This can be found using Newton's second law:

F space equals space m a

Worked example

An electron travels at right angles to a uniform magnetic field of flux density 6.2 mT. The speed of the electron is 3.0 × 106 m s-1 and its charge-to-mass ratio is 1.8 × 1011 C kg-1.

Calculate the radius of the circular path of the electron.

Answer:

Step 1: Write down the known quantities

  • Charge-to-mass ratio: q over m space equals space 1.8 space cross times space 10 to the power of 11 space straight C space kg to the power of negative 1 end exponent
  • 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

r space equals fraction numerator space m v over denominator B q end fraction

Step 3: Substitute in values

m over q space equals fraction numerator space 1 over denominator 1.8 space cross times space 10 to the power of 11 end fraction

r space equals space fraction numerator open parentheses 3.0 space cross times space 10 to the power of 6 close parentheses space over denominator open parentheses 1.8 space cross times space 10 to the power of 11 close parentheses open parentheses 6.2 space cross times space 10 to the power of negative 3 end exponent close parentheses end fraction space equals space 2.688 space cross times space 10 to the power of negative 3 end exponent space straight m space equals space 2.7 space mm thin space open parentheses 2 space straight s. straight f close parentheses 

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Ashika

Author: Ashika

Expertise: Physics Project Lead

Ashika graduated with a first-class Physics degree from Manchester University and, having worked as a software engineer, focused on Physics education, creating engaging content to help students across all levels. Now an experienced GCSE and A Level Physics and Maths tutor, Ashika helps to grow and improve our Physics resources.