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Wize University Physics Textbook (Master) > Magnetic Fields and Magnetic Force

Velocity Selectors

Charged Particle Motion in a Magnetic FieldPrevious Section
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Velocity Selectors


If a charged particle moves through a magnetic field, it can be made to travel in a straight line if there is just the right amount of electric field present as well.
  • A charged particle moving through a magnetic and electric field will experience two forces:
  • The electric force: ∣Felec∣=qE|F_{elec}|=qE
  • The magnetic force: ∣Fmag∣=qvBsin⁡θ|F_{mag}|=qvB\sin\theta
  • By Newton's Second Law, if the net force is zero, then there is zero acceleration, and there cannot be circular motion. The two forces must be equal and opposite:
ΣF⃗=ma⃗F⃗elec+F⃗mag=(0)∣F⃗elec∣=∣F⃗mag∣\begin{aligned} \Sigma \vec F=m\vec a\\ \vec F_{elec}+\vec F_{mag}=(0)\\ |\vec F_{elec}|=|\vec F_{mag}|\\ \end{aligned}
  • We can take advantage of this to find the magnitude of the two fields required for particles to move in a straight line:
∣F⃗elec∣=∣F⃗mag∣qE=qvBE=vBv=EB\begin{aligned} |\vec F_{elec}|&=|\vec F_{mag}|\\ qE &= qvB \\ E &= vB \\ v &= \frac{E}{B} \end{aligned}
  • These setups are sometimes called velocity selectors because any particles not moving at this exact speed will not travel in a straight line.
  • This effectively filters particles travelling with a certain speed.
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Example: Proton Travelling in a Straight Line


A proton is travelling along a straight line and enters a region where both electric and magnetic fields are present. It is observed that proton does not change its line of motion and moves in a straight line.

If the initial velocity of the proton is 6.0×106 m/s to the right, and the magnitude of electric field is 6000 N/C, what is the magnitude of the magnetic field?


Since the proton travels in a straight line without changing its speed, there must be zero acceleration on the proton. By Newton's 2nd Law, if there is zero acceleration, there must be zero net force - so the electric and magnetic forces must be equal and opposite (we are ignoring gravity).

Setting the electric and magnetic forces equal to each other:
∣F⃗elec∣=∣F⃗mag∣qE=qvBsin⁡θqE=qvBsin⁡(90o)E=vBB=EvB=6×103 N/C6×106 m/sB=1.0 mT\begin{aligned} |\vec F_{elec|}&= |\vec F_{mag}| \\ qE&= qvB\sin\theta \\ qE&= qvB\sin(90^o) \\ E&= vB \\ B&=\frac{E}{v}\\ B&=\frac{6\times10^3~N/C}{6\times10^6~m/s}\\ B&=1.0 ~mT \end{aligned}

Practice: Electron Travelling in a Straight Line


Consider an electron moving to the right through uniform electric and magnetic fields. The electric field points up the page.

a) Find the direction of the magnetic field.
b) Provided that the electric field strength is 90 kV/m90 ~kV/m and the magnetic field strength is 0.5 T0.5~ T, find the speed of the electron such that the electron continues to travel in a straight line along the x-direction.


Direction of magnetic field in left figure