Answer:
B = 4.45mT in the +^k direction
Explanation:
In order to calculate the required magnitude of the magnetic force, to achieve that the beam of particles emerge undeflected of the parallel plates, the electric force between the plates and the magnetic field in that region must be equal.
[tex]F_E=F_B\\\\qE=qvB[/tex] (1)
q: charge of the particles beam = +2e = 2*1.6*10^-19C
v: speed of the particles = ?
B: magnitude of the magnetic field = ?
E: electric field between the plates = V/d
V: potential difference between the parallel plates = 150V
d: distance of separation of the plates = 0.00820m
If you assume that the below plate is negative, the electric force on the particles has a direction upward (+^j). Then, the direction of the magnetic force must be downwards (-^j).
To obtain a downward magnetic force, it is necessary that the magnetic field point out of the page. In fact, if the direction of motion of the particles is toward east (+^i) and the magnetic field points out of the page (+^k), you have:
^i X ^k = -^j
Furthermore, it is necessary to calculate the sped of the particles. It is made by using the information about the charge, the potential difference that accelerates the particles and the kinetic energy.
[tex]K=qV=\frac{1}{2}mv^2\\\\v=\sqrt{\frac{2qV}{m}}[/tex] (2)
You replace the expression (2) into the equation (1) and solve for B:
[tex]B=\frac{E}{v}=E\sqrt{\frac{m}{2qV}}[/tex]
[tex]B=\frac{V}{d}\sqrt{\frac{m}{2qV}}\\\\B=\frac{150V}{0.0820m}\sqrt{\frac{6.64*10^{-27}kg}{2(2(1,6*10^{-19}C))(1.75*10^3V)}}\\\\B=4.45*10^{-3}T=4.45mT[/tex]
The required magnitude of the magnetic field is 4.45mT and has a direction out of the page +^k
Following are the solution to the given question:
In order to emerge using reflecting means, use the following formula:
[tex]\to F_E = F_B ..............(1)\\\\ \to F_E = \text{electric force}\\\\ \to F_B = \text{magnetic force}\\\\[/tex]
Calculating the Lorent's force:
[tex]\to F=qE+qv \times B \ \ also,\ \ K_{E} =U_{E} \\\\[/tex]
[tex]\to K_{E}[/tex][tex]= \text{kinetic energy} = -\frac{1}{2} \ mv^2 \\\\[/tex]
[tex]\to U_{E} = \text{potential energy} = q_V[/tex]
Calculating the value of v: \\\\
[tex]\to v= \sqrt{\frac{2qV}{m}} \\\\ \to q = 2e^{+} = 2 (1.6 \times 10^{-19}\ C) = 3.2 \times 10^{-19} C \\\\\to V = 1.75 \times 10^{3} \V \\\\\to m = 6.64 \times 10^{-27} \ kg\\\\ \to v = 410,700.25 \ \frac{m}{s}\\\\[/tex]
It's the particle's velocity, but the velocity also is defined as:
[tex]\to v=\frac{E}{B}[/tex]
solving for B:
[tex]\to B= \frac{E}{v}\\\\[/tex]
[tex]= \frac{\frac{V}{d}}{v}\\\\ =\frac{V}{vd} \\\\= \frac{150\ V}{(410,700.25 \ \frac{m}{s}) (8.2 \times 10^{-3} m)} \\\\= 0.045\ T\\\\[/tex]
When indicated in the diagram, the direction is parallel to "v" and E.
Learn more:
brainly.com/question/15522069
