F = BQv sinθ
The force on an isolated moving charge is perpendicular to its motion and the magnetic field B
F = BQv
The electron experiences a force upwards when it travels through the magnetic field between the two poles
An electron is moving at 5.3 × 107 m s-1 in a uniform magnetic field of flux density 0.2 T.Calculate the force on the electron when it is moving at 30° to the field, and state the factor it increases by compared to when it travels perpendicular to the field.
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
A charged particle moves travels in a circular path in a magnetic field
An electron with charge-to-mass ratio of 1.8 × 1011 C kg-1 is travelling 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.Calculate the radius of the circle path of the electron.
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
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