# Current Induced in a Wire

1. Jun 29, 2013

If a wire moves with some constant velocity (4 m/s for example in the diagram) parallel to the direction of the wire itself (unlike in most physics problems in which you drag a wire 'sideways' or perpendicular to the direction of the wire) but still perpendicular to the magnetic field, what is the direction of the current produced?

We know that the force created by the current needs to oppose the direction of motion, so by the right hand rule, it would seem that the case illustrated in the diagram would yield a current in the 'up' direction, as in there would be electrons moving 'down' across the diameter of the wire. Would that mean that there is a voltage across the diameter of the wire? or am I missing something entirely?

#### Attached Files:

• ###### Wire.jpg
File size:
111.7 KB
Views:
126
2. Jun 30, 2013

### tiny-tim

Physics is equations.

What is the equation for this?

3. Jun 30, 2013

Tiny-tim, this is at least the second time you've slyly guided people to the right answer.

Alright, I would use

$\epsilon$$=$$B$$l$$\nu$
or maybe better...
$\epsilon$$=\frac{-\Delta(BAcos(\theta))}{\Delta t}$

So since obviously you don't have a changing area, and you're moving already inside a magnetic field, you're not going to get a current since the magnetic field isn't changing. Stupid of me, but I suppose I meant more of a diagram like the one below in which the wire is moving into a magnetic field.

So, I guess the more fundamental question I mean to ask is How accurately can we treat a wire as an object that has cross sectional area, and what implications does that hold for voltage and current?

With the square wire set shown, the first equation stated is great, and from the voltage and resistance you can calculate the current which makes things easier..

#### Attached Files:

File size:
23.5 KB
Views:
125
• ###### Wire Box.jpg
File size:
17 KB
Views:
120
4. Jun 30, 2013

### tiny-tim

he he!
A wire certainly does have a width, albeit a tiny one that will have only a tiny potential difference.

But don't forget, it won't have a current unless there's a circuit

which there won't be unless you somehow join the top to the bottom "the long way round".