# Force on a wire carrying current

1. Jun 9, 2012

### ehabmozart

It is given in my book. hat F=B I l sin theta where theta is the angle between B and the current I... I guess this is no right.. I mean, how would the current make an angle with B, it should b the length of wire itself... Secondly F=Bqv is the reason behind a charged particle which enter perpendicularly to a magnetic field moves in a circle.. My doubt is that at point, this point charge will be inline with the field line.. From where will it get force at that point.. Talking about the quarter circle... I need more clarification and it would be more than amazing if there are some illustrative diagrams!

2. Jun 9, 2012

### MalachiK

The angle in your first equation IS the angle between the current and the field lines. If you imagine a uniform B field you can also imagine a wire in this field pointing in various different directions.

Current is a vector, and only the component of the current vector that cuts across the field lines causes the force. When the current is at right angles to the field, θ = 90 and sin θ =1. This means that all of the current contributes to the force so you get the maximum force = BIL.

On the other hand, when the current is parallel to the field, θ = 0 and sin θ = 0. This makes your force = BIL × 0 = 0 as there is no component of the current cutting across the field lines.