Baking bread with yeast is it chemical physical changes
2 answers:
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A)
The strength of the magnetic field produced by a current-carrying wire is given by
where
is the vacuum permeability
I is the current in the wire
r is the distance from the wire
In this situation,
I = 5.20 A
r = 4.60 cm = 0.046 m is the distance of the electron from the wire
Therefore the magnetic field strength at the electron's location is
The force exerted on a charged moving particle travelling perpendicular to a magnetic field is given by
where
q is the magnitude of the charge of the particle
v is its velocity
B is the magnetic flux density
For the electron of this problem,
is the charge
is the speed
is the magnetic field
Substituting,
B) Same direction as the current in the wire
First of all we have to find the direction of the magnetic field lines, which are concentric around the wire. Assuming the wire carries a current pointing upward, then if we use the right-hand rule:
- The thumb gives the direction of the current -> upward
- The other fingers wrapped give the direction of the field lines -> anticlockwise around the wire (as seen from top)
Now the direction of the force can be found by using the right-hand rule. We have:
- direction of the index finger = direction of motion of the electron (toward the wire, let's assume from east to west)
- middle finger = direction of the magnetic field (to the north)
- Thumb = direction of the force --> downward
However, the electron carries a negative charge, so we must reverse the direction of the force: therefore, the force experienced by the electron will be upward, so in the same direction as the current in the wire.