Current in wire in Magnetic field[Easy QUICK question]

[orangefruit]

Homework Statement

Suppose a very long straight wire were held perpendicular to the plane of the page such that it intersects the page at point D. Additionally, suppose the wire was being moved toward the top of the page.

Suppose instead that the wire were moved toward the left side of the page. Would the current through the wire flow into the page, out of the page, or not at all?

Homework Equations

Right hand rules. F=qv x B. F= IL x B.

The Attempt at a Solution

I think this is into the page. I used qVxB. My index finger is going to the left. Middle finger straight. So the force is down. The force pushes positive currents into the page. Is this correct? My other way was F= ILxB. But it gave me out of the page. I think it's because I treat the force as the v. But I think that's wrong because that F is magnetic. Thank you for the help. This is a little urgent

 

[orangefruit]

Anyone? :)

 

[jfgobin]

Good Morning OrangeFruit,

In order for a current to be generated in a conductor due to of magnetic field, what are the conditions?

 

[deep838]

My other way was F= ILxB. But it gave me out of the page. I think it's because I treat the force as the v. But I think that's wrong because that F is magnetic. Thank you for the help. This is a little urgent

both of these point into the page... F=q*vxB where q is positive... but the current is due to negative charges so here the equation is F=-q*vxB=-ILxB...

 

[Nickie]

.

Your explanation is correct. When the wire is moved towards the left side of the page, the current through the wire will flow into the page. This can be determined using the right hand rule for magnetic forces (F=qv x B). The force is directed downwards, so the current will flow into the page. It is important to remember that the force in this case is the force exerted by the magnetic field on the moving charges in the wire, not the force on the wire itself. Using the equation F=ILxB would give the incorrect result because the force in this case is not the same as the force experienced by the wire.