Physics: An electron of kinetic energy 1.67 kev circles in a plane perpendicular to a uniform magnetic field. the orbit radius is 37.4 cm. find (a) the electron s speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

18joules Explanation:1/2 of 4=2 2 multiply by 9= 18 joules...

An electron of kinetic energy 1.67 kev circles

hannahrasco4051

08.01.2022

Explanation:

Given that,

The kinetic energy of the electron, $K=1.67\ keV=1.67\times 10^3\ eV$

Radius of the orbit, r = 37.4 cm = 0.374 m

To find,

(a) the electron's speed, (b) the magnetic field magnitude, (c) the circling frequency, and (d) the period of the motion.

Solve,

We know that, $1\ eV=1.6\times 10^{-19}\ J$

$K=1.67\times 10^3\times 1.6\times 10^{-19}\ J=2.67\times 10^{-16}\ J$

(a) The formula of the kinetic energy is given by :

$K=\dfrac{1}{2}mv^2$

v is the speed if electron

m is the mass of electron

$v=\sqrt{\dfrac{2K}{m}}$

$v=\sqrt{\dfrac{2\times 2.67\times 10^{-16}}{9.1\times 10^{-31}}}$

$v=2.42\times 10^7\ m/s$

(b) Let B is the magnitude of magnetic field. On the circular path the magnetic field is given by :

$B=\dfrac{mv}{qr}$

$B=\dfrac{9.1\times 10^{-31}\times 2.42\times 10^7}{1.6\times 10^{-19}\times 0.374}$

$B=3.68\times 10^{-4}\ T$

$T=\dfrac{2\pi r}{v}$

Circling frequency is given by :

$f=\dfrac{1}{T}=\dfrac{v}{2\pi r}$

$f=\dfrac{2.42\times 10^7}{2\pi \times 0.374 }$

$f=1.02\times 10^7\ Hz$

(d) The period of motion,

$T=\dfrac{1}{f}$

$T=\dfrac{1}{1.02\times 10^7}$

$T=9.80\times 10^{-8}\ s$

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