Question

FIGURE 1 shows part of a mass spectrometer. The whole arrangement is in a vacuum. Negative ions of mass 2.84 x 10-20 kg and charge -2.0 x 10-19 C are produced at S, which is at a potential difference of - 3000 V. The ions are accelerated in a narrow beam towards H, a hole in a hollow metal container. Inside the container, the negative ions enter a region where an electric field of strength E and a magnetic field of strength 0.83 T act.

Is it possible for the io
ns to exit slit P without being deflected in the fields?
Explain, with the aid of diagram.​

Answer 1

Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C

Explanation:

When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:

• The force due to the electric field, whose magnitude is $F_E=qE$, where q is the charge of the ion and E is the strength of the electric field
• The force due to the magnetic field, whose magnitude is $F_B=qvB$, where v is the speed of the ions and B is the strength of the magnetic field

The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if

$F_E = F_B$

which translates into

$qE=qvB\\\rightarrow v = \frac{E}{B}$

Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.

We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference $V=-3000 V$, so we have that their kinetic energy is given by the change in electric potential energy:

$qV=\frac{1}{2}mv^2$

where

$q=-2.0\cdot 10^{-19}C\\m=2.84\cdot 10^{-20}kg$

Solving for v, the speed,

$v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-2.0\cdot 10^{-19})(-3000)}{2.84\cdot 10^{-20}}}=205.6 m/s$

And since the magnetic field strength is

B = 0.83 T

The strength of the electric field must be

$E=vB=(205.6 m/s)(0.83 T)=170.6 N/C$

Learn more about electric and magnetic fields:

brainly.com/question/8960054

brainly.com/question/4273177

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly