An electron is moving northward in a magnetic field. The magnetic force on the electron is toward the northeast. What is the dir
1 answer:
To solve this problem we will use the vector concept given by the cross product between two perpendicular vectors and which results in a vector perpendicular to these two. From the definition of the Magnetic Force we have to
From the property of cross product the magnetic force should point in the direction perpendicular to the plane containing the vectors v and B.
The direction of velocity is north, and the direction of the magnetic force is northeast.
This cannot be the case, as the direction of magnetic force is not perpendicular to the direction of velocity of the charge.
Therefore the correct option for the direction of the magnetic field is <em>"This situation cannot exist because of the relative orientations of the velocity and force vectors" </em>
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Answer:
(a) 1.87 x 10⁻⁴ m/s
(b) 0.013V
Explanation:
(a) Drift speed, , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;
=
----------------(i)
Where;
q = amount of charge
n = free charge density
A = cross-sectional area of the wire
But current density, J, is the electric current per unit cross-section area. This is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e
J = =
Equation (i) can then be written as follows;
=
=
=
---------------------(ii)
From the question;
E = 0.0520N/C
p = 1.72 x 10⁻⁸ Ωm
n = 8.5 x 10²⁸ electrons/m³
c = charge on electron = 1.9 x 10⁻¹⁹C
Substitute these values into equation (ii) as follows;
=
= 1.87 x 10⁻⁴ m/s
(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e
V = E x d [where d = 25.0cm = 0.25m]
V = 0.0520 x 0.25
V = 0.013V