Finite wire in magnetic field, determine time and electric potential

[scholio]

Homework Statement

a wire 0.3 meters in length is suspended in a region in which a uniform magnetic field of 4Teslas points into the page as shown ( see attach.). the wire is dropped at time t = 0 seconds. at what times will there be an electric potential of 98 volts between the ends of the wire? also, which end of the wire will have the higher potential?

Homework Equations

position x = 1/2(at^2) where a is acceleration due to gravity = 9.8m/s/s, t is time

force of gravity F = mg where m is mass, g is gravity = 9.8m/s/s

1/2(mv^2) = qV where q is charge, V is electric potential

magnetic force F_m = qv X B where q is charge, v is velocity, B is magnetic field, X indicates cross product

gravitational potential energy U = mgh where m is mass, g is gravity = 9.8 m/s/s, h is height = x

gravitational potential energy = kinetic energy = 1/2 (CV^2) = 1/2(mv^2) where C is capacitance, V is electric potential, m is mass, v is velocity

The Attempt at a Solution

i'm guessing most of the above equations are not needed at all.

i'm not too sure how to do this, but this is what I've been scheming: use the " 1/2(mv^2) = qV " to solve for v while holding m and q constant.

sub in v into magnetic force eq and solve for force, and since force F = ma = m ( 1/2(at^2)) where m is constant, and a = 9.8m/s/s i can solve for t.

is this close? i haven't included the given length of 0.3 meters, though.

as for the second 'part', determining which end of the wire has the higher potential i am not too sure about, tips appreciated.

 

[rl.bhat]

In relevant equations one important equation is missing. And in this problem where the capacitance comes into the picture?
Can you write down Farady's law of electromagnetic induction?

 

[scholio]

faraday's lay of induction:

epsilon = - d(magnetic flux)/dt where epsilon is the induced emf in the circuit, dt is change in time and d(magnetic flux) is change in magnetic flux, where magnetic flux = [integral(B dA)] where B is magnetic field, dA is change in area.

capacitance, since capacitance C = Q/V, and i don't have C nor Q, i don't think it actually applies in this problem

 

[Niles]

Hmm, can't this be done simpler?

When the wire moves down the magnetic field, the magnetic force will make the positive charges pile up in the right side of the wire. hence we will have an electric field going from right to left.

This electric force will at some point equal the magnetic force, and hence qE = qvB, which gives us E = vB.

Now we know that V = EL, so V = vBL. Since we know the direction of the electric field, we know which side has the higher potential.

EDIT: Faraday's law of induction only applies to closed circuits, which is not the case here.

 

[scholio]

how do you know that positive charges pile up on the right side of the wire?

isn't the magnetic force qv X B = qvBsin(theta), not qvB or are you assuming theta = 90 degrees?

so since E = vB ---> V = EL (which equation is this?) --> V = vBL where V = 98 volts, B = 4 Teslas, L = 0.3 meters, then v = 81.67 m/s

and since v_f = v_i + at where v_f = v, v_i = 0, a = 9.8 m/s/s, then t = 8.33 seconds

and since positive charge builds up on right side of wire, and electric field moves right to left, then the side with higher potential is the right side.

is my time correct? did i determine the correct side with higher potential?

 

[Niles]

1) Yes, theta is 90 degrees, so F = qvB. I use the right-hand rule to determine the direction.

2) The potential V = EL I find using

$$V = - \int {E \cdot dl}$$

3) I get the same time and direction.