Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude 2607 N/C, directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of 6.81 103 m/s, enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of +3.72 10-12 C enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is 1.80 10-9 N, pointing directly upward. What is the speed of this particle?
To solve this problem, we can use the principle that a charged particle moving through both electric and magnetic fields will experience a force described by the Lorentz force equation.
The Lorentz force equation is given by F = q(E + v x B), where F is the net force acting on the particle, q is the charge of the particle, E is the electric field, v is the velocity of the particle, and B is the magnetic field.
In this problem, we are given that the electric field magnitude is 2607 N/C, directed vertically upward, the magnetic field is directed south, the charge of the particle is +3.72 x 10^-12 C, and the net force acting on the particle is 1.80 x 10^-9 N, pointing directly upward. The particle is also traveling east at a speed of 6.81 x 10^3 m/s.
To find the speed of the particle, we need to rearrange the Lorentz force equation to solve for the magnitude of the velocity. The equation becomes:
F = q(E + v x B)
F = qE + q(v x B)
Since the force is directed upward and the electric field is directed upward, the force due to the electric field is qE, and the force due to the vector cross product term is q(v x B). Therefore, we can write the equation as:
1.80 x 10^-9 N = 3.72 x 10^-12 C x 2607 N/C + 3.72 x 10^-12 C x (v x B)
Now, we need to express the vector cross product term in terms of magnitudes. The magnitude of the vector cross product is given by the product of the magnitudes of the vectors and the sine of the angle between the vectors. In this case, the angle between the velocity vector (east) and the magnetic field vector (south) is 90 degrees.
The magnitude of the velocity vector is 6.81 x 10^3 m/s, and the magnitude of the magnetic field is not given. Let's denote it as B_mag.
Therefore, we can write the equation as:
1.80 x 10^-9 N = 3.72 x 10^-12 C x 2607 N/C + 3.72 x 10^-12 C x (6.81 x 10^3 m/s)(B_mag)(sin 90°)
Since sin 90° is equal to 1, the equation simplifies to:
1.80 x 10^-9 N = 3.72 x 10^-12 C x 2607 N/C + 3.72 x 10^-12 C x (6.81 x 10^3 m/s)(B_mag)
Now, we can solve for B_mag by isolating it on one side of the equation:
3.72 x 10^-12 C x (6.81 x 10^3 m/s)(B_mag) = 1.80 x 10^-9 N - 3.72 x 10^-12 C x 2607 N/C
Simplifying further:
B_mag = (1.80 x 10^-9 N - 3.72 x 10^-12 C x 2607 N/C)/(3.72 x 10^-12 C x 6.81 x 10^3 m/s)
Finally, we can substitute the given values into the equation and calculate the magnitude of the magnetic field (B_mag).
Motion of the particle without deflection =>
F(el) = F mag)
qE =qvB
B=E/v=2607/6.81•10³=0.38 T
For the second particle
F=qE-qv₁B
v₁=(F+qE)/qB=
=1.8•10⁻⁹•3.72•10⁻¹²•2607}/
3.72•10⁻¹²•0.38 =
=8.13•10³ m/s