# Question on the Lorentz force: Why is the force not F=q(v×B) = F=qv×qB

• I
unplebeian
TL;DR Summary
Why is the charge not multiplied to the cross product
Background: is the equation of Lorentz force for the force acting on a moving charge in electric and magnetic field.

For the magnetic field only it is : F=qv×B.

Question:
For magnetic field only why is the force not F=q(v×B) = F=qv×qB

Last edited by a moderator:

Gold Member
MHB
TL;DR Summary: Why is the charge not multiplied to the cross product

Background: is the equation of Lorentz force for the force acting on a moving charge in electric and magnetic field.

For the magnetic field only it is : F=qv×B.

Question:
For magnetic field only why is the force not F=q(v×B) = F=qv×qB
You are only multiplying by q once, so
##q \textbf{v} \times \textbf{B}##

##= q ( \textbf{v} \times \textbf{B} )##

## = (q \textbf{v} ) \times \textbf{B}##

##= \textbf{v} \times (q \textbf{B})##

-Dan

• Ibix
unplebeian
Hi, Dan,
I'm sorry I didn't get it. That is a scalar multiplication so q should be multiplied to both. Generally a(bxc)= abxac.
Why are we multiplying only once?

Homework Helper
Gold Member
2022 Award
Generally a(bxc)= abxac.
This is wrong.
$$a(\mathbf b \times \mathbf c) = a\mathbf b \times \mathbf c = \mathbf b \times a\mathbf c$$You must be thinking of:
$$a(\mathbf b + \mathbf c) = a\mathbf b + a\mathbf c$$

• Gold Member
MHB
Hi, Dan,
I'm sorry I didn't get it. That is a scalar multiplication so q should be multiplied to both. Generally a(bxc)= abxac.
Why are we multiplying only once?
Is ##2(3 \times 4 ) = (2 \cdot 3) \times (2 \cdot 4)##?

-Dan

• This is wrong.
$$a(\mathbf b \times \mathbf c) = a\mathbf b \times \mathbf c = \mathbf b \times a\mathbf c$$You must be thinking of:
$$a(\mathbf b + \mathbf c) = a\mathbf b + a\mathbf c$$
Easy to make mistake if in elementary school you learned the order of operations as "Dot (##\cdot## and ##\colon##) before stroke (##+## and ##-##)", because that's how the basic operators are written in your country.

Last edited:
• topsquark
unplebeian
Thank you, Dan. I thought about it graphically and it's evident that the scalar multiplication to both vectors prior to the cross product operation is incorrect. Rather take the cross product and then perform the scalar multiplication or simply any one vector like you suggested.

Thank you.

• berkeman