Question

A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.

(a) What must the speed of a particle be for it to pass through undeflected?
________m/s
(b)What energy must protons have to pass through undeflected?
________keV
(c) What energy must electrons have to pass through undeflected?
________eV

Answer 1

Answer:

speed of a particle  = 2.31 × $10^{6}$ m/s

energy of proton required = 27.77 KeV

energy of electron required =  15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and  energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB    ..............1

and force due to electric field E = qE   .....................2

so without deflection force due to magnetic field  = force due to electric field  

so here qvB = qE

and V = $\frac{E}{B}$    ...................3

put here value

V =  $\frac{0.51*10^6}{0.22}$

speed of a particle  = 2.31 × $10^{6}$ m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × $\frac{V^2}{2}$

put here value

energy of proton required = 1.67 × $10^{-27}$ × $\frac{(2.31*10^6)^2}{2}$

energy of proton required = 4.45 × $10^{-15}$ J

energy of proton required = 4.45 × $10^{-15}$ J ÷ (1.602 × $10^{-19}$

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × $\frac{V^2}{2}$

put here value

energy of electron required = 9.11 × $10^{-31}$ × $\frac{(2.31*10^6)^2}{2}$

energy of electron required =  24.305× $10^{-19}$ J

energy of electron required =  24.305 × $10^{-19}$ J ÷ (1.602 × $10^{-19}$  

energy of electron required =  15.171 eV

Answer 2

(a) The speed of the particle must be 2.32×10⁶ m/s

(b) The kinetic energy of the proton must be 27.93 J

(c) The kinetic energy of the electron must be 15.3 eV

Electric and magnetic field:

(a) The force on a charged particle due to Electric field is given by:

F = qE

where q is the charge on the particle, and

E is the electric field.

The magnetic force on a charged particle moving perpendicular to the magnetic field is given by:

F' = qvB

where v is the velocity of the particle, and

B is the magnetic field.

Far a particle to pass undeflected in a region where electric field and magnetic field are perpendicular to each other will be:

qE =  qvB

$v=\frac{E}{B}$

$v=\frac{0.51\times10^6}{0.22}\\\\v=2.32\times10^6\;m/s$

(b) The kinetic energy required by proton to pass undeflected will be:

$KE_p=\frac{1}{2}m_pv^2$

where $m_p$ is the mass of proton = $1.67\times10^{-27}kg$

so,

$KE_p=0.5\times1.67\times10^{-27}\times(2.32\times10^6)^2\;J\\\\KE_p=4.47\times10^{-15}\;J$

converting to eV:

$KE_p=\frac{4.47\times10^{-15}}{1.6\times10^{-19}}=27.93\;keV$

(c) The kinetic energy required by electron to pass undeflected will be:

$KE_e=\frac{1}{2}m_ev^2$

where $m_p$ is the mass of electron = $9.1\times10^{-31}kg$

so,

$KE_p=0.5\times9.1\times10^{-31}\times(2.32\times10^6)^2\;J\\\\KE_p=2.45\times10^{-18}\;J$

converting to eV:

$KE_p=\frac{2.45\times10^{-18}}{1.6\times10^{-19}}=27.93\;keV=15.3\;eV$

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