**Lorentz Force**

The electromagnetic field gives rise to a Lorentz force that acts on electrically charged particles:

**F**= q**E** + q**v **x **B**

where bold means vector representation: **F** is the force on the charge q, **E** is the electric field, **v** is the velocity of the charge and **B** is the magnetic field.

According to the equation, the Lorentz force has two contributions. One, the magnetic, is the cross product of the velocity and the magnetic field. This causes the resultant force to be perpendicular to both the velocity and the magnetic field. The other contribution, the electric one, is parallel to the electric field.

**Charged particle moving through a magnetic field**

A charged particle moving through a magnetic field of strength $B$ with a speed $v$ will feel a Lorentz force (**Figure 1**) with a magnitude of:

$F=qvBsinθ$

where $θ$ is the angle between the velocity of the particle and the magnetic field (field lines run from N to S) and $q$ is the charge of the particle. This force acts at right angles to both the magnetic field and the velocity of the particle. Fleming’s left-hand rule (where the thumb represents force, the first finger the magnetic field, and the second finger the velocity if the particle is positively charged) can be used to remember the direction of this force. In terms of co-ordinates, if velocity is aligned with the $+x$ axis and the field aligned with $+y$, then the force is in the $+z$ direction.

**Current-carrying wire in a magnetic field**

A current-carrying wire in a magnetic field will feel a Lorentz force in a direction given by Fleming’s left hand rule, with a magnitude of:

$F=IlBsinθ$

where $l$ is the length of the wire in the magnetic field, $I$ is the current flowing through the wire and $θ$ is the angle between the wire and the magnetic field.

Read also: Electrodynamics | Explanation, Formulas, Questions and Answers

**Questions and Answers – Lorentz force**

**What is the magnitude of the force experienced when the unit charge is kept under the influence of 5 N/C electric fields?**

Answer:

The for on the charge is given by,

F = qE

⇒ F = (1)(5)

⇒ F = 5 N/s.

**If the magnitude of the force experienced when a -2C charge is moving at 10m/s under the influence of 5 N/C electric fields. The magnetic field of 5 magnitudes is 30° to the direction of the electric field and velocity. What is the magnitude of the force experienced by the charge?**

Answer:

The for on the charge is given by,

F = qE + q(v × B)

⇒ F = (-2)(5) + (-2) (10 × 5 × sin(30))

⇒ F = -10 + -100 × 0.5

⇒ F = -60 N

Sources: PinterPandai, Coursera, Xabier Cid Vidal & Ramon Cid

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